Definition7.1. The impulse response of (3.1) is the derivative of the step response.
Remark7.2. The step response may not be differentiable as a function at
time zero and the derivative in Definition 7.1 should be understood in the sense of distributions. When \(D\neq 0\), then the step response is not even continuous at zero and the impulse response is a
distribution rather than a function.
For general \(D\), the impulse response equals the above function plus \(D\delta \) where \(\delta \) is the Dirac delta (which you may have encountered in other modules) which is the distributional derivative of the unit step function. We will
never explicitly need the case \(D\neq 0\) though.
An interpretation of the impulse response is that it is the output for the input \(\delta \) (which in engineering jargon is called an “impulse”).
Remark7.3. We note that by definition the impulse response is uniquely determined by the step response. In the
other direction, we have that the step response is uniquely determined by the impulse response. When \(D=0\), the step response \(H\) can be obtained from the impulse response \(h\) though solving the initial value problem \(H'=h\),
\(H(0)=0\). When \(D\neq 0\) something similar is still true, but making this precise requires some knowledge of distributions which we do not want to go into.
Remark7.4. The transfer function is the Laplace transform of the impulse response and the frequency response is
the Fourier transform of the impulse response. By inverse Laplace or Fourier transforming, we can obtain the impulse response from the transfer function or frequency response.
simply states that for \(x^0=0\), the output equals the convolution of the input with the impulse response. In particular, the impulse response uniquely determines the output (for \(x^0=0\)) for any input. Since any of the step response, transfer
function and frequency response uniquely determine the impulse response, any of these three also uniquely determines the output (for \(x^0=0\)) for any input. This justifies the earlier Remark 4.4.
Remark7.6. If \(m=1\) (i.e. \(B\) is a vector) and \(D=0\), then the solution
of the initial value problem
\[ \dot {x}=Ax,\qquad x(0)=B,\qquad y=Cx, \]
is \(y(t)=C\e ^{At}B\) and therefore equals the impulse response.
(a) Impulse response for second order system (low pass).
(b) Impulse response for second order system (band pass).
Figure 7.2: Impulse responses for second order systems.
7.2 Case study: a suspension system
We consider the fixed structure suspension system. In Figure 7.3 we give the impulse response from the road velocity to each of the three performance outputs. Here we have chosen
values for the various physical parameters as in the introduction. The road velocity being the Dirac delta correponds to the road profile being the Heaviside step function. This corresponds for example to mounting a curb.
From Figure 7.3 we see that the outputs are oscillatory (as expected) and the amplitude of the oscillations is significant for about two seconds.
Figure 7.3: Impulse responses for the three outputs \(z_1\), \(z_2\) and \(z_3\) for reasonable (non-optimal) values of \(d\) and \(k_s\).
7.3 Problems
Consider the second order (low pass) system
\[ \ddot {y}+4\dot {y}+3y=3u. \]
(a) Write this in standard form \(\dot {x}=Ax+Bu\), \(y=Cx\) with \(x:=\sbm {y\\\dot {y}}\).
(c) Calculate the step response by solving \(\ddot {y}+4\dot {y}+3y=3\), \(y(0)=\dot {y}(0)=0\) and then differentiate this to obtain the impulse response.
Solution. A particular solution is \(y=1\). The general solution of the homogeneous equation is \(\ddot {y}+4\dot {y}+3y=0\). It characteristic polynomial is \(s^2+4s+3=(s+3)(s+1)\). So the general
solution of the homogeneous problem is \(a\e ^{-3t}+b\e ^{-t}\). So the general solution of the original problem is \(y(t)=1+a\e ^{-3t}+b\e ^{-t}\). The initial conditions give \(1+a+b=0\), \(-3a-b=0\). Solving this gives \(a=\frac {1}{2}\),
\(b=-\frac {3}{2}\). So the step response is
The general solution of the differential equation is \(y(t)=a\e ^{-3t}+b\e ^{-t}\). The initial conditions give \(a+b=0\), \(-3a-b=3\). Solving this gives \(a=-\frac {3}{2}\), \(b=\frac {3}{2}\). Therefore we obtain
