Linear Constant Coefficient Differential Equations

Introduction

This is a step-by-step guide for finding the impulse response for Linear Constant Coefficient Differential Equations (LCCDE). Once you find the impulse response, you can find the response to any other input signal using continuous convolution. It is based on Linear Constant Coefficient Discrete Difference Equations and was created by asking Gemini to create a continuous version of that page but to focus on solving the step response first and then taking the derivative of the step response to get the impulse response.

Very Basic First Derivative

Let's start with the simplest case:

$ \frac{dy(t)}{dt}+a_0\,y(t)=x(t) $

Rather than solving for the impulse response directly with a singular delta function, we will first find the step response $ s_r(t) $—the output when the input is a unit step, $ x(t)=u(t) $. Once we have the step response, we can find the impulse response $ h(t) $ by taking the time derivative:

$ h(t) = \frac{d s_r(t)}{dt} $

Step Response $ s_r(t) $

To find the step response, we solve:

$ \frac{ds_r(t)}{dt}+a_0\,s_r(t)=u(t) $

Assuming the system is initially at rest (causal LTI), $ s_r(t)=0 $ for all $ t<0 $. Because the input $ u(t) $ is bounded, the state cannot jump instantaneously, meaning:

$ s_r(0^+) = s_r(0^-) = 0 $

For $ t>0 $, the input is constant ($ u(t)=1 $), and the total solution consists of a homogeneous part plus a particular (forced) part:

$ s_r(t) = y_h(t) + y_p(t), \quad t>0 $

• Particular Solution: Since the forcing function is a constant ($ 1 $), we guess a constant $ y_p(t)=K $:

$ \frac{dK}{dt}+a_0\,K=1 \implies 0+a_0\,K=1 \implies y_p(t) = \frac{1}{a_0} $

• Homogeneous Solution: Replacing the right side with 0 yields the characteristic equation:

$ s + a_0 = 0 \implies s = -a_0 \implies y_h(t) = C\,e^{-a_0 t} $

Combining these gives:

$ s_r(t) = \left(C\,e^{-a_0 t} + \frac{1}{a_0}\right)u(t) $

We determine $ C $ using our initial condition at $ t=0^+ $:

$ s_r(0^+) = C\,e^{0} + \frac{1}{a_0} = 0 \implies C = -\frac{1}{a_0} $

Factoring out the constant, the step response is:

$ s_r(t) = \frac{1}{a_0}\left(1 - e^{-a_0 t}\right)u(t) $

Impulse Response $ h(t) $

Since $ \delta(t) = \frac{du(t)}{dt} $, linearity and time-invariance dictate that:

$ h(t) = \frac{ds_r(t)}{dt} $

Applying the product rule across the discontinuity at $ t=0 $:

$ \begin{align*} h(t) &= \frac{d}{dt}\left[\frac{1}{a_0}\left(1 - e^{-a_0 t}\right)\right]u(t) + \left[\frac{1}{a_0}\left(1 - e^{-a_0 t}\right)\right]\frac{du(t)}{dt}\\ &= \frac{1}{a_0}\left(0 - (-a_0)e^{-a_0 t}\right)u(t) + \left[\frac{1}{a_0}\left(1 - e^{-a_0 t}\right)\right]\delta(t) \end{align*} $

Using the sifting property of the impulse ($ f(t)\delta(t) = f(0)\delta(t) $):

$ \left[\frac{1}{a_0}\left(1 - e^{-a_0 (0)}\right)\right]\delta(t) = \frac{1}{a_0}(1 - 1)\delta(t) = 0 $

The impulse term vanishes completely, leaving:

$ h(t) = e^{-a_0 t}\,u(t) $

Slightly Less Basic First Derivative

Now let's include a non-unity coefficient $ a_1 $ on the derivative term:

$ a_1\,\frac{dy(t)}{dt}+a_0\,y(t)=x(t) $

We start by normalizing the equation by $ a_1 $:

$ \frac{dy(t)}{dt}+\frac{a_0}{a_1}\,y(t)=\frac{1}{a_1}x(t) $

Just as before, we first solve for the step response $ s_r(t) $ when the input is $ x(t)=u(t) $, then differentiate to get the impulse response $ h(t) $.

Step Response $ s_r(t) $

To find the step response, we solve:

$ \frac{ds_r(t)}{dt}+\frac{a_0}{a_1}\,s_r(t)=\frac{1}{a_1}u(t) $

Assuming the system is initially at rest (causal LTI), $ s_r(t)=0 $ for all $ t<0 $. With a bounded input $ u(t) $, the state variable cannot jump instantaneously, giving the initial condition:

$ s_r(0^+) = s_r(0^-) = 0 $

For $ t>0 $, the forcing function is constant ($ u(t)=1 $), so the total solution is the sum of the homogeneous and particular parts:

$ s_r(t) = y_h(t) + y_p(t), \quad t>0 $

• Particular Solution: Since the right-hand side is constant ($ \frac{1}{a_1} $), we guess a constant $ y_p(t)=K $:

$ \frac{dK}{dt}+\frac{a_0}{a_1}\,K=\frac{1}{a_1} \implies 0 + \frac{a_0}{a_1}\,K = \frac{1}{a_1} \implies y_p(t) = \frac{1}{a_0} $

• Homogeneous Solution: Replacing the right side with 0 gives the characteristic equation:

$ s + \frac{a_0}{a_1} = 0 \implies s = -\frac{a_0}{a_1} \implies y_h(t) = C\,e^{-\frac{a_0}{a_1}t} $

Combining these gives:

$ s_r(t) = \left(C\,e^{-\frac{a_0}{a_1}t} + \frac{1}{a_0}\right)u(t) $

We determine $ C $ using the initial condition at $ t=0^+ $:

$ s_r(0^+) = C\,e^0 + \frac{1}{a_0} = 0 \implies C = -\frac{1}{a_0} $

Factoring out $ \frac{1}{a_0} $, the step response is:

$ s_r(t) = \frac{1}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)u(t) $

Impulse Response $ h(t) $

Now we find $ h(t) $ by differentiating the step response with respect to time:

$ h(t) = \frac{ds_r(t)}{dt} $

Using the product rule across the jump at $ t=0 $:

$ \begin{align*} h(t) &= \frac{d}{dt}\left[\frac{1}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)\right]u(t) + \left[\frac{1}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)\right]\frac{du(t)}{dt} \\ &= \frac{1}{a_0}\left(0 - \left(-\frac{a_0}{a_1}\right)e^{-\frac{a_0}{a_1}t}\right)u(t) + \left[\frac{1}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)\right]\delta(t) \end{align*} $

Using the sifting property on the impulse term ($ f(t)\delta(t) = f(0)\delta(t) $):

$ \left[\frac{1}{a_0}\left(1 - e^{-\frac{a_0}{a_1}(0)}\right)\right]\delta(t) = \frac{1}{a_0}(1 - 1)\delta(t) = 0 $

The delta term drops out completely, leaving:

$ h(t) = \frac{1}{a_1}\,e^{-\frac{a_0}{a_1}t}\,u(t) $

General Homogeneous Solution

For LCCDE problems, we use a general homogeneous trial solution of $ y_h(t)=C\,e^{st} $, where $ s $ is a constant based on the coefficients.

When finding the step response for a system with only an unshifted input term on the right side:

$ a_1\,\frac{dy(t)}{dt}+a_0\,y(t)=b_0\,x(t) $

we normalize by $ a_1 $:

$ \frac{dy(t)}{dt}+\frac{a_0}{a_1}\,y(t)=\frac{b_0}{a_1}x(t) $

To find the step response $ s_r(t) $, set $ x(t)=u(t) $. Because the system starts from rest, $ s_r(t)=0 $ for $ t<0 $, and the state variable cannot jump instantaneously at $ t=0 $:

$ s_r(0^+) = s_r(0^-) = 0 $

For $ t>0 $, the forcing function is constant ($ u(t)=1 $), giving a total response of:

$ s_r(t) = y_h(t) + y_p(t), \quad t>0 $

• Particular Solution: Guessing a constant $ y_p(t) = K $:

$ 0 + \frac{a_0}{a_1}\,K = \frac{b_0}{a_1} \implies y_p(t) = \frac{b_0}{a_0} $

• Homogeneous Solution: Substituting $ y_h(t)=C\,e^{st} $ into the unforced equation yields:

$ \begin{align*} C\,s\,e^{st} + \frac{a_0}{a_1}\,C\,e^{st} &= 0\\ C\,e^{st}\left(s+\frac{a_0}{a_1}\right) &= 0 \end{align*} $

For a non-trivial solution ($ C \neq 0 $), we must have $ s = -\frac{a_0}{a_1} $.

Combining these for $ t>0 $:

$ s_r(t) = C\,e^{-\frac{a_0}{a_1}t} + \frac{b_0}{a_0} $

Using the initial rest condition $ s_r(0^+) = 0 $:

$ s_r(0^+) = C + \frac{b_0}{a_0} = 0 \implies C = -\frac{b_0}{a_0} $

This gives the step response:

$ s_r(t) = \frac{b_0}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)u(t) $

Taking the time derivative to find the impulse response $ h(t) = \frac{ds_r(t)}{dt} $:

$ \begin{align*} h(t) &= \frac{d}{dt}\left[\frac{b_0}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)\right]u(t) + \left[\frac{b_0}{a_0}\left(1 - e^{-\frac{a_0}{a_1}t}\right)\right]\delta(t) \\ &= \frac{b_0}{a_0}\left(0 - \left(-\frac{a_0}{a_1}\right)e^{-\frac{a_0}{a_1}t}\right)u(t) + 0 \\ &= \frac{b_0}{a_1}\,e^{-\frac{a_0}{a_1}t}\,u(t) \end{align*} $

Characteristic Polynomial

From this point forward, for first- and higher-order systems we can systematically determine the exponential rates $ s $ by setting the forcing function to zero and replacing each $ k $th derivative $ \frac{d^k y(t)}{dt^k} $ with $ s^k $.

For example, if we have:

$ \frac{d^2y(t)}{dt^2}+6\,\frac{dy(t)}{dt}+8\,y(t)=x(t) $

the characteristic polynomial is:

$ s^2+6s+8=0 $

which factors as $ (s+2)(s+4)=0 $, indicating that $ s=-2,-4 $.

Example 1: Fits Basic Case

Imagine you want to solve for the impulse response of:

$ 3\,\frac{dy(t)}{dt}+2\,y(t)=x(t) $

• Normalize the equation based on the coefficient of the highest derivative:

$ \frac{dy(t)}{dt}+\frac{2}{3}\,y(t)=\frac{1}{3}x(t) $

• Find $ s $ by replacing $ x(t) $ with 0 and all $ \frac{d^k y}{dt^k} $ with $ s^k $:

$ \begin{align*}s+\frac{2}{3}&=0\\ s&=-\frac{2}{3}\end{align*} $

• Find the step response $ s_r(t) $ by setting $ x(t)=1 $ for $ t>0 $:
  • The particular solution is a constant: $ 0 + \frac{2}{3}K = \frac{1}{3} \implies y_p(t) = \frac{1}{2} $
  • The total response for $ t>0 $ is $ s_r(t) = C\,e^{-\frac{2}{3}t} + \frac{1}{2} $
  • Using initial rest $ s_r(0^+) = 0 $:

$ s_r(0^+) = C + \frac{1}{2} = 0 \implies C = -\frac{1}{2} $

• • Thus:

$ s_r(t) = \frac{1}{2}\left(1 - e^{-\frac{2}{3}t}\right)u(t) $

• Differentiate $ s_r(t) $ to find the impulse response $ h(t) $:

$ \begin{align*} h(t) &= \frac{ds_r(t)}{dt} = \frac{d}{dt}\left[\frac{1}{2}\left(1 - e^{-\frac{2}{3}t}\right)\right]u(t) + \underbrace{\left[\frac{1}{2}\left(1 - e^0\right)\right]\delta(t)}_{0} \\ h(t) &= \frac{1}{3}\,e^{-\frac{2}{3}t}\,u(t) \end{align*} $

Example 2: Fits Basic Case After Scaling and Rearranging

Imagine you want to solve for the impulse response of:

$ 6\,\frac{dy(t)}{dt}=2\,x(t)-4\,y(t) $

• Adjust the equation so that all $ y $ terms are on the left, all $ x $ terms are on the right, and the highest derivative of $ y $ has a coefficient of 1:

$ \begin{align*} 6\,\frac{dy(t)}{dt}+4\,y(t)&=2\,x(t)\\ \frac{dy(t)}{dt}+\frac{2}{3}\,y(t)&=\frac{1}{3}x(t) \end{align*} $

• At this point, this matches Example 1 above.
• Solving for the step response first:

$ s_r(t) = \frac{1}{2}\left(1 - e^{-\frac{2}{3}t}\right)u(t) $

• Differentiating yields the impulse response:

$ h(t) = \frac{ds_r(t)}{dt} = \frac{1}{3}\,e^{-\frac{2}{3}t}\,u(t) $

More Complicated Forcing Function

All the cases above could be re-written to have a single $ x(t) $ on the right. If the right-hand-side ends up more complicated than $ x(t) $ by including derivatives of the input, it turns out that you can still solve the problem by re-writing things such that the left side contains a new variable $ z(t) $, solving as if the right side at that point were merely $ x(t) $, and then relating $ z(t) $ to the more complicated right side using scaling and derivative operators.

Reminder: Convolutions with Impulses and Derivatives

Recall that:

$ f(t) \ast a\,\delta(t) = a\,f(t), \qquad f(t) \ast a\,\delta'(t) = a\,\frac{df(t)}{dt} $

That means that something like $ \frac{3}{2}\,x(t)-\frac{7}{2}\,\frac{dx(t)}{dt} $ could also be written as $ x(t)\ast\left(\frac{3}{2}\,\delta(t)-\frac{7}{2}\,\delta'(t)\right) $.

Back to Basics

Let's start with:

$ 2\,\frac{dy(t)}{dt}+5\,y(t)=3\,x(t)-7\,\frac{dx(t)}{dt} $

The issue now is the complicated right side—even after we normalize by 2 to get the coefficient of $ \frac{dy(t)}{dt} $ to be 1, there will be more than just $ x(t) $ on the right. So let's start by normalizing the equation based on $ a_1 $:

$ \frac{dy(t)}{dt}+\frac{5}{2}\,y(t)=\frac{3}{2}x(t)-\frac{7}{2}\,\frac{dx(t)}{dt} $

Now I am going to solve a slightly different equation—one where the only variable on the right is $ x(t) $. I am going to replace the variables on the left with a new variable, $ z(t) $:

$ \frac{dz(t)}{dt}+\frac{5}{2}\,z(t)=x(t) $

Note that if I convolve both sides of this equation with $ \frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t) $ I get:

$ \begin{align*} \left(\frac{dz(t)}{dt}+\frac{5}{2}\,z(t)\right)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right)&=x(t)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right) \end{align*} $

If I distribute the convolution on the right I get:

$ \begin{align*} \left(\frac{dz(t)}{dt}+\frac{5}{2}\,z(t)\right)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right)&=\frac{3}{2}x(t)-\frac{7}{2}\,\frac{dx(t)}{dt} \end{align*} $

which is the original right side of my equation; I can then replace it with the original left side of my equation:

$ \begin{align*} \left(\frac{dz(t)}{dt}+\frac{5}{2}\,z(t)\right)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right)&=\frac{dy(t)}{dt}+\frac{5}{2}\,y(t) \end{align*} $

Now I can replace derivatives with un-differentiated signals convolved with derivatives of deltas:

$ \begin{align*} z(t)\ast\left(\delta'(t)+\frac{5}{2}\,\delta(t)\right)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right)&= y(t)\ast\left(\delta'(t)+\frac{5}{2}\,\delta(t)\right) \end{align*} $

I will rearrange the order of the convolution on the left side a bit to get:

$ \begin{align*} z(t)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right)\ast\left(\delta'(t)+\frac{5}{2}\,\delta(t)\right)&= y(t)\ast\left(\delta'(t)+\frac{5}{2}\,\delta(t)\right) \end{align*} $

For these equations to be equal, $ y(t)=z(t)\ast\left(\frac{3}{2}\delta(t)-\frac{7}{2}\,\delta'(t)\right)=\frac{3}{2}\,z(t)-\frac{7}{2}\,\frac{dz(t)}{dt} $. This means if I can solve for the impulse response $ h_z(t) $ for the $ z(t) $ equation, I can find the impulse response $ h_y(t)=\frac{3}{2}\,h_z(t)-\frac{7}{2}\,\frac{dh_z(t)}{dt} $. We know the general solution for a first-derivative problem already, so can quickly find:

$ h_z(t)=e^{-\frac{5}{2}t}\,u(t) $

Taking the derivative of $ h_z(t) $ using the product rule:

$ \begin{align*} \frac{dh_z(t)}{dt} &= \frac{d}{dt}\left[e^{-\frac{5}{2}t}\right]u(t) + e^{-\frac{5}{2}t}\frac{du(t)}{dt} \\ &= -\frac{5}{2}e^{-\frac{5}{2}t}u(t) + e^{-\frac{5}{2}(0)}\delta(t) \\ &= -\frac{5}{2}e^{-\frac{5}{2}t}u(t) + \delta(t) \end{align*} $

and from that we can assemble $ h_y(t) $:

$ \begin{align*} h_y(t)&=\frac{3}{2}h_z(t)-\frac{7}{2}\,\frac{dh_z(t)}{dt}\\ &=\frac{3}{2}e^{-\frac{5}{2}t}\,u(t) - \frac{7}{2}\left(-\frac{5}{2}e^{-\frac{5}{2}t}u(t) + \delta(t)\right)\\ &=-\frac{7}{2}\delta(t) + \left(\frac{3}{2} + \frac{35}{4}\right)e^{-\frac{5}{2}t}\,u(t)\\ &=-\frac{7}{2}\delta(t) + \frac{41}{4}e^{-\frac{5}{2}t}\,u(t) \end{align*} $

Another First-Derivative Example

Find the impulse response $ h_y(t) $ if the right side consists solely of an input derivative:

$ \begin{align*} \frac{dy(t)}{dt}+\frac{1}{2}\,y(t)&=3\,\frac{dx(t)}{dt} \end{align*} $

• First, solve for the simplified impulse response $ h_z(t) $ assuming just an unscaled, undifferentiated $ x(t) $ on the right:

$ \begin{align*} \frac{dz(t)}{dt}+\frac{1}{2}\,z(t)&=x(t)\\ h_z(t)&=e^{-\frac{1}{2}t}\,u(t) \end{align*} $

• Using the linearity of the system, relate $ h_y(t) $ directly to the derivative operator on the right:

$ h_y(t)=3\,\frac{dh_z(t)}{dt} $

• Differentiate $ h_z(t) $ using the product rule:

$ \begin{align*} \frac{dh_z(t)}{dt} &= \frac{d}{dt}\left[e^{-\frac{1}{2}t}\right]u(t) + e^{-\frac{1}{2}t}\frac{du(t)}{dt}\\ &= -\frac{1}{2}e^{-\frac{1}{2}t}u(t) + e^{-\frac{1}{2}(0)}\delta(t)\\ &= -\frac{1}{2}e^{-\frac{1}{2}t}u(t) + \delta(t) \end{align*} $

• Scaling by 3 gives the final impulse response:

$ \begin{align*} h_y(t)&=3\,\delta(t)-\frac{3}{2}e^{-\frac{1}{2}t}\,u(t) \end{align*} $

Big Picture

For a continuous differential equation of the general form:

$ \begin{align*} \sum_{k=0}^{N}a_k\,\frac{d^k y(t)}{dt^k}&=\sum_{l=0}^{M}b_l\,\frac{d^l x(t)}{dt^l} \end{align*} $

the process breaks down into three standard steps:

1. Find the characteristic roots $ s $: Set the right side to 0 and replace the $ k $th derivative with $ s^k $:$ \begin{align*} \sum_{k=0}^{N}a_k\,s^k&=0 \end{align*} $
2. Solve for the simplified step response $ s_{r,z}(t) $ and impulse response $ h_z(t) $: Replace the left-hand dependent variable with $ z(t) $ and simplify the right side to just $ x(t) $:$ \begin{align*} \sum_{k=0}^{N}a_k\,\frac{d^k z(t)}{dt^k}&=x(t) \end{align*} $
   • Find the step response $ s_{r,z}(t) $ for $ x(t)=u(t) $ using the particular constant solution and homogeneous modes, enforcing rest initial conditions at $ t=0^+ $ ($ s_{r,z}(0^+)=0, s_{r,z}'(0^+)=0, \dots $).
   • Take the time derivative of $ s_{r,z}(t) $ to get the simplified impulse response:$ h_z(t) = \frac{d s_{r,z}(t)}{dt} $
3. Construct the full impulse response $ h_y(t) $: Apply the input differential operator directly to $ h_z(t) $: $ \begin{align*} h_y(t)&=\sum_{l=0}^{M}b_l\,\frac{d^l h_z(t)}{dt^l} \end{align*} $

Second Derivative and More

If the left side of the equation has higher-order derivatives than just $ \frac{dy(t)}{dt}+a_0\,y(t) $, the nature of the characteristic roots $ s $ expands from a simple real constant $ s=-a_0 $ to all the possible solutions for an $ N $th-order polynomial, where $ N $ is the highest derivative on the left side of the equation. Here are the possibilities for $ N $th-order polynomials (assuming real-valued coefficients):

• For a 1st order polynomial, $ s $ will be a single real value.
• For a 2nd order polynomial, $ s $ could be:
  • two different real values,
  • a real value repeated twice, or
  • a complex conjugate pair ($ \sigma \pm j\omega $)
• For a 3rd order polynomial, $ s $ could be a real value and then any of the possibilities for 2nd order polynomials, so:
  • three different real values,
  • one real value and a different real value repeated twice,
  • a real value repeated three times, or
  • a real value and a complex conjugate pair
• For a 4th order polynomial, $ s $ could be any of the possibilities for a 2nd order polynomial combined with any of the possibilities for a 2nd order polynomial, so:
  • All real
    • four different real values,
    • two different real values and a third real value repeated twice,
    • two different real values, each repeated twice,
    • two different real values, one repeated three times,
    • one real value repeated four times,
  • One pair of complex conjugates plus two real roots
    • two different real values and a complex conjugate pair,
    • one real value repeated twice and a complex conjugate pair,
  • Two pairs of complex conjugates
    • two different complex conjugate pairs, or
    • a repeated set of complex conjugate pairs
• After this, the pattern continues with some combination of (possibly repeated) real values and (possibly repeated) complex conjugate pairs.

Each type of root $ s $ defines the functional form of the homogeneous response:

• Individual real roots $ s $ have $ C\,e^{st}u(t) $
• Real roots $ s $ repeated $ p $ times have a $ (p-1) $th order polynomial along with the $ e^{st} $ term, meaning $ \sum_{k=0}^{p-1}C_k\,t^k\,e^{st}u(t) $
• Complex conjugate pairs $ s=\sigma \pm j\omega $ create sinusoids $ C\,e^{\sigma t}\cos(\omega t+\phi)u(t) $ or $ e^{\sigma t}\left(C_c\cos(\omega t)+C_s\sin(\omega t)\right)u(t) $
• Complex conjugate pairs $ s=\sigma \pm j\omega $ repeated $ p $ times have a $ (p-1) $th order polynomial along with the sinusoids meaning $ \sum_{k=0}^{p-1}C_k\,t^k\,e^{\sigma t}\cos(\omega t+\phi_k)u(t) $ or $ \sum_{k=0}^{p-1}t^k\,e^{\sigma t}\left(C_{c,k}\cos(\omega t)+C_{s,k}\sin(\omega t)\right)u(t) $

Initial Rest Conditions for the Step Response $ s_{r,z}(t) $

When solving for the simplified step response of an $ N $th-order system:

$ \frac{d^N z(t)}{dt^N} + a_{N-1}\frac{d^{N-1}z(t)}{dt^{N-1}} + \dots + a_0 z(t) = u(t) $

the input $ u(t) $ is bounded (equal to 1 for $ t>0 $). Because finite forcing cannot produce instantaneous jumps across integrations, the state and its first $ N-1 $ derivatives must all be continuous at $ t=0 $.

Assuming the system starts from initial rest, every initial condition evaluated at $ t=0^+ $ is simply zero:

$ \begin{align*} s_{r,z}(0^+) &= 0\\ s_{r,z}'(0^+) &= 0\\ s_{r,z}''(0^+) &= 0\\ &\;\;\vdots\\ s_{r,z}^{(N-1)}(0^+) &= 0 \end{align*} $

This makes determining the unknown coefficients $ C_i $ straightforward:

1. Find the constant particular solution $ y_p(t) = \frac{1}{a_0} $ (assuming $ a_0 \neq 0 $).
2. Set up the total step response for $ t>0 $: $ s_{r,z}(t) = y_h(t) + y_p(t) $.
3. Solve the system of $ N $ linear equations created by setting $ s_{r,z}(0^+)=0, \dots, s_{r,z}^{(N-1)}(0^+)=0 $.
4. Differentiate $ s_{r,z}(t) $ once with respect to time to get $ h_z(t) $.

Second-Derivative Example: Distinct Real Roots

Find the impulse response $ h_y(t) $ for:

$ \begin{align*} \frac{d^2y(t)}{dt^2}+6\,\frac{dy(t)}{dt}+8\,y(t)&=x(t) \end{align*} $

• First, ensure all $ y $ terms are on the left, the highest derivative has a coefficient of 1, and the right side is simply $ x(t) $ — already done!
• Since $ x(t) $ is unscaled and has no derivatives on the right, we do not need the intermediate $ z(t) $ step.
• Find the characteristic roots by replacing $ \frac{d^k y}{dt^k} $ with $ s^k $ and setting the forcing function to 0:

$ \begin{align*} s^2+6s+8&=0\\ (s+2)(s+4)&=0 \implies s = -2, -4 \end{align*} $

• This means the homogeneous response for $ t>0 $ is:

$ y_h(t) = C_1\,e^{-2t} + C_2\,e^{-4t} $

Step Response $ s_r(t) $

We find the step response by setting $ x(t)=u(t) $. For $ t>0 $, $ x(t)=1 $.

• Particular Solution: Guessing a constant $ y_p(t) = K $:

$ 0 + 6(0) + 8K = 1 \implies y_p(t) = \frac{1}{8} $

• The total step response for $ t>0 $ is:

$ s_r(t) = C_1\,e^{-2t} + C_2\,e^{-4t} + \frac{1}{8} $

• Because the system is initially at rest and $ u(t) $ is finite, the initial conditions at $ t=0^+ $ are both zero:

$ s_r(0^+) = 0, \qquad s_r'(0^+) = 0 $

• Applying $ s_r(0^+) = 0 $:

$ C_1 + C_2 + \frac{1}{8} = 0 \implies C_1 + C_2 = -\frac{1}{8} $

• Taking the derivative of $ s_r(t) $ for $ t>0 $:

$ s_r'(t) = -2C_1\,e^{-2t} - 4C_2\,e^{-4t} $

• Applying $ s_r'(0^+) = 0 $:

$ -2C_1 - 4C_2 = 0 \implies C_1 = -2C_2 $

• Substituting $ C_1 = -2C_2 $ into the first equation:

$ -2C_2 + C_2 = -\frac{1}{8} \implies -C_2 = -\frac{1}{8} \implies C_2 = \frac{1}{8} $

• Solving for $ C_1 $:

$ C_1 = -2\left(\frac{1}{8}\right) = -\frac{1}{4} $

• Thus, the step response is:

$ s_r(t) = \left(-\frac{1}{4}e^{-2t} + \frac{1}{8}e^{-4t} + \frac{1}{8}\right)u(t) = \frac{1}{8}\left(1 - 2e^{-2t} + e^{-4t}\right)u(t) $

Impulse Response $ h_y(t) $

Now we find $ h_y(t) $ by taking the time derivative of $ s_r(t) $:

$ h_y(t) = \frac{ds_r(t)}{dt} $

Because $ s_r(0^+) = 0 $, no impulse term appears at $ t=0 $ ($ s_r(0^+)\delta(t) = 0 $). Differentiating the terms for $ t>0 $:

$ \begin{align*} h_y(t) &= \left[-\frac{1}{4}(-2)e^{-2t} + \frac{1}{8}(-4)e^{-4t} + 0\right]u(t)\\ &= \left(\frac{1}{2}e^{-2t} - \frac{1}{2}e^{-4t}\right)u(t) \end{align*} $

Factoring out $ \frac{1}{2} $:

$ \begin{align*} h_y(t) &= \frac{1}{2}\left(e^{-2t} - e^{-4t}\right)u(t) \end{align*} $

Second Derivative: Complex $ s $

Find the impulse response for:

$ \frac{d^2y(t)}{dt^2}-\frac{dy(t)}{dt}+1.69\,y(t)=x(t) $

• First, ensure all $ y $ terms are on the left, the highest derivative has a coefficient of 1, and the right side is simply $ x(t) $ — already there!
• Since $ x(t) $ has no derivatives on the right, we do not need the intermediate $ z(t) $ step.
• Replace $ \frac{d^k y}{dt^k} $ with $ s^k $ and replace $ x(t) $ with 0 to find characteristic roots:

$ \begin{align*} s^2-s+1.69&=0\\ s&=\frac{1\pm\sqrt{1-6.76}}{2}=\frac{1\pm j2.4}{2}=0.5 \pm j1.2 \end{align*} $

• Here, the real growth rate is $ \sigma=0.5 $ and the oscillation frequency is $ \omega=1.2\text{ rad/s} $.
• This means the homogeneous solution for $ t>0 $ takes the form:

$ y_h(t) = e^{0.5t}\left(C_1\cos(1.2t)+C_2\sin(1.2t)\right) $

Step Response $ s_r(t) $

We find the step response by setting $ x(t)=u(t) $. For $ t>0 $, $ x(t)=1 $.

• Particular Solution: Guessing a constant $ y_p(t) = K $:

$ 0 - 0 + 1.69K = 1 \implies y_p(t) = \frac{1}{1.69} = \frac{100}{169} $

• The total step response for $ t>0 $ is:

$ s_r(t) = e^{0.5t}\left(C_1\cos(1.2t)+C_2\sin(1.2t)\right) + \frac{100}{169} $

• Because the system is initially at rest and $ u(t) $ is finite, the initial conditions at $ t=0^+ $ are both zero:

$ s_r(0^+) = 0, \qquad s_r'(0^+) = 0 $

• Applying $ s_r(0^+) = 0 $:

$ \begin{align*} s_r(0^+) &= e^0\left(C_1\cos(0) + C_2\sin(0)\right) + \frac{100}{169} = 0\\ C_1 + \frac{100}{169} &= 0 \implies C_1 = -\frac{100}{169} \end{align*} $

• Taking the derivative of $ s_r(t) $ for $ t>0 $ using the product rule:

$ \begin{align*} s_r'(t) &= 0.5\,e^{0.5t}\left(C_1\cos(1.2t)+C_2\sin(1.2t)\right) + e^{0.5t}\left(-1.2\,C_1\sin(1.2t)+1.2\,C_2\cos(1.2t)\right) \end{align*} $

• Applying $ s_r'(0^+) = 0 $:

$ \begin{align*} s_r'(0^+) &= 0.5\left(C_1(1) + 0\right) + \left(0 + 1.2\,C_2(1)\right) = 0\\ 0.5\,C_1 + 1.2\,C_2 &= 0\\ 1.2\,C_2 &= -0.5\,C_1\\ C_2 &= -\frac{0.5}{1.2}\,C_1 = -\frac{5}{12}\,C_1 \end{align*} $

• Substituting $ C_1 = -\frac{100}{169} $:

$ C_2 = -\frac{5}{12}\left(-\frac{100}{169}\right) = \frac{125}{507} $

• Thus, the step response is:

$ s_r(t) = \left[e^{0.5t}\left(-\frac{100}{169}\cos(1.2t) + \frac{125}{507}\sin(1.2t)\right) + \frac{100}{169}\right]u(t) $

Impulse Response $ h_y(t) $

Now we find $ h_y(t) $ by taking the time derivative of $ s_r(t) $:

$ h_y(t) = \frac{ds_r(t)}{dt} $

Because $ s_r(0^+) = 0 $, the step jump across $ t=0 $ is zero ($ s_r(0^+)\delta(t) = 0 $), so no impulse term appears.

Differentiating for $ t>0 $ (which is simply $ s_r'(t) $):

$ \begin{align*} h_y(t) &= e^{0.5t}\left[\left(0.5\,C_1 + 1.2\,C_2\right)\cos(1.2t) + \left(0.5\,C_2 - 1.2\,C_1\right)\sin(1.2t)\right]u(t) \end{align*} $

Notice how cleanly the terms simplify using our earlier equations:

• The cosine coefficient is $ 0.5\,C_1 + 1.2\,C_2 = s_r'(0^+) = 0 $.
• The sine coefficient evaluates directly: $ \begin{align*} 0.5\,C_2 - 1.2\,C_1 &= 0.5\left(\frac{125}{507}\right) - 1.2\left(-\frac{100}{169}\right)\\ &= \frac{125}{1014} + \frac{120}{169} = \frac{125}{1014} + \frac{720}{1014} = \frac{845}{1014} = \frac{5}{6} \end{align*} $
• (Notice that $ 845 = 5 \times 169 $ and $ 1014 = 6 \times 169 $).*

Assembling the final causal impulse response:

$ \begin{align*} h_y(t)&=\frac{5}{6}\,e^{0.5t}\sin(1.2t)\,u(t) \end{align*} $