Frequency response of LTI systems
We have seen that unit complex exponentials are the basic building blocks of frequency domain descriptions of signals. In this chapter we investigate how an LTI system responds when the input is a unit complex exponential at a given frequency ω. We will see that the output is a unit complex exponential at the same frequency ω, but scaled by a complex number. That complex number is called the frequency response at frequency ω of the system. In this Chapter we define the frequency response of CT and DT LTI systems. We discuss the relationship with the impulse response, and show how to find the frequency response of a variety of systems.
6.1 Frequency response
6.1.1 Response of LTI systems to complex exponentials
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CT frequency response
Suppose S is a BIBO stable continuous-time LTI system with input x and output y = S(x). If x(t) = ejωt = cos(ωt) + j sin(ωt) is a complex exponential with frequency ω then there is a complex number H(ω) such that
y(t) = H(ω)ejωt = H(ω)x(t).
The complex number H(ω) is called the frequency response at frequency ω. The function H is called the frequency response of the system.
One way to see that this this is true is to use the convolution formula. If h is the impulse response of the system and the input is x(t) = ejωt then
∞ ∞ y(t) = h(τ)x(t − τ) dτ =
jω(t−τ) jωt ∞ −jωτ
dτ = e h(τ)e dτ. (6.1)
Notice that we could move the ejωt outside of the integral because the variable of integration is τ not t. From the BIBO stability of the system, we know that the output must be bounded, and that the integral on the right hand-side of (6.1) converges. If we define
∞ −jωτ h(τ)e dτ
y(t) = H(ω)ejωt.
Relationship between CT frequency response and impulse response
For a BIBO stable continuous-time LTI system, the frequency response at frequency ω and the impulse response h are related by
h(τ)e dτ. (6.3)
We discuss integrals of this form in Topic 8 which deals with the continuous-time Fourier transform.
DT frequency response
Suppose S is a BIBO stable discrete-time LTI system with input x and output y = S(x). If x(t) = ejωn is a complex exponential with frequency ω then there is a complex number H(ω) such that
y[n] = H(ω)ejωn = H(ω)x[n].
The complex number H(ω) is called the frequency response at frequency ω. The function H is called the frequency response of the system.
The basic argument we used to show this in the continuous-time case is still valid in the discrete- time case. All we need to do is use the DT convolution formula instead of the CT convolution formula. In this case, if the impulse response of the system is h, then the frequency response is related to the impulse response as follows.
Relationship between DT frequency response and impulse response
For a BIBO stable discrete-time LTI system, the frequency response at frequency ω and the impulse response h are related by
H(ω) = h[k]e−jωk (6.4)
We discuss sums of this form in Topic 8, which covers the discrete-time Fourier transform.
6.1.2 Magnitude and phase response
The frequency response at frequency ω is generally a complex number. For the sake of inter- pretation, we usually think of this compex number in polar form as
H(ω) = |H(ω)|ej∠H(ω).
The real number |H(ω)| is the magnitude response at frequency ω and the real number ∠H(ω)
is the phase response at frequency ω.
Real systems and magnitude and phase response
Systems of physical interest usually have the property that if the input signal is real for all time, then the output signal is also real for all time. We call these real systems. For real systems, the impulse response h is a real signal. The frequency response of real systems has special symmetry properties.
Symmetry properties of frequency response for real systems
If S is a real LTI system then H(−ω) = H(ω)∗. This means that
|H(−ω)| = |H(ω)∗| = |H(ω)| i.e., the magnitude response is even ∠H(−ω) = ∠(H(ω)∗) = −∠H(ω) i.e., the phase response is odd.
If a CT LTI system is real, then h(τ)∗ = h(τ) for all τ. Using this observation we can see that for any ω,
∗∞ −jωτ∗∞ ∗−jωτ∗ ∞ −j(−ω)τ H(ω) = h(τ)e dτ = h(τ) (e ) dτ = h(τ)e
dτ = H(−ω).
A similar argument holds for a real DT LTI system. In that case the impulse response is real and so satisfies h[k]∗ = h[k] for all k. Then, using (6.4), we can show that H(ω)∗ = H(−ω) for all ω.
Response of LTI system to a sinusoid
One reason it is useful to think about the frequency response in terms of the magnitude and phase response is that this helps us understand the output of an LTI system when the input is cos(ωt), a sinusoid at a given frequency.
Suppose S is a real system, so its frequency response satisfies |H(−ω)| = |H(ω)| and ∠H(−ω) = −∠H(ω). If the input is
then the output is
x(t) = cos(ωt) = 1 ejωt + e−jωt 2
y(t) = 21 H(ω)ejωt + H(−ω)e−jωt
= |H(ω)|21 ej(ωt+∠H(ω)) + e−j(ωt+∠H(ω))
= |H(ω)| cos(ωt + ∠H(ω)).
So the magnitude response scales the amplitude of the sinusoid, and the phase response causes a phase shift.
6.1.3 DT frequency response is periodic
For the discrete-time unit complex exponential
ejωn = ej(ω+2π)n for all integers n and all real ω.
This is because ej2πn = 1 for all n. A consequence of this fact is that the DT frequency response must be periodic with period 2π.
DT frequency response is periodic with period 2π
If H is the frequency response of a DT LTI system then H(ω) = H(ω + 2π) for all ω.
To see why, suppose the input to a DT LTI system is ejωn. Then, by the definition of the frequency response, the output is H(ω)ejωn. If the input to a DT LTI system is ej(ω+2π)n then the output is H(ω + 2π)ej(ω+2π)n. But since ejωn = ej(ω+2π)n for all n and all ω, we must have H(ω) = H(ω + 2π) for all ω.
Summary of Section 6.1
If the input to a BIBO stable LTI system is a complex exponential with frequency ω, the output is a complex exponential at frequency ω scaled by a complex number H(ω) depending on the frequency. The function H is called the frequency response of the system. The frequency response of an LTI system is closely related to the impulse response of the system and completely characterises the system.
6.2 Finding the frequency response
In this section we find the frequency response of a number of systems by explicitly finding the system output when the input is x(t) = ejωt (in the CT case) or x[t] = ejωn in the DT case.
Example 6.1
Consider the CT time-shifting system Delayτ . Recall that the output of this system is the input delayed by τ time units. If x(t) = ejωt is the input then the output is
y(t) = x(t − τ) = ejω(t−τ) = e−jωτ ejωt. So the frequency response of Delayτ is
H(ω) = e−jωτ .
The magnitude response is |H(ω)| = |e−jωτ | = 1 for all ω. The phase response is
∠H(ω) = −ωτ for all ω.
Note that the phase response is a linear function for delay systems.
Example 6.2
Consider the CT differentiator system. If the input is x(t) = ejωt then the output is y(t) = dx = jωejωt.
So the frequency response of the CT differentiator is H(ω) = jω for all ω.
The magnitude response is |H(ω)| = |ω| for all ω. The phase response is π/2 if ω > 0
∠H(ω)= −π/2 ifω<0.
Note that the phase response is undefined at ω = 0 since the frequency response is zero at ω = 0.
Example 6.3
Consider the DT unit delay system Delay1. Recall that the output of this system is the output delayed by one time unit. If x[n] = ejωn is the input then the output is
y[n] = x[n − 1] = ejω(n−1) = e−jωxjωn. So the frequency response of the DT unit delay is
H(ω) = e−jω.
The magnitude response is |H(ω)| = 1 for all ω. The phase response is
∠H(ω) = −ω for all ω.
Example 6.4
Consider the DT three-point averaging system. For this system, the output can be ex- pressed in terms of the input according to the following difference equation:
y[n] = 13 (x[n] + x[n − 1] + x[n − 2]) for all n. If x[n] = ejωn then the output is
1 jωn jω(n−1) y[n]=3e +e
jω(n−2) 1 −jω =31+e +e
−2jω jωn e .
So the frequency response of the three-point averaging system is
H(ω)= 11+e−jω +e−2jω. 3
The magnitude and phase response of this system are shown below:
ω −2π −π 0 π 2π
−2π −π 0 π 2πω − π3
We can also find formulas for the magnitude and phase response. To do this, we first
rewrite the frequency response as
e−jω jω −jω −jω 2cos(ω)+1 e + 1 + e = e .
|2 cos(ω) + 1| −ω if cos(ω) > −1/2 3 and ∠H(ω) = π − ω if cos(ω) < −1/2.
Note that there are many different ways to write the phase (because ej(φ+2π) = ejφ for all φ). Our expression, for instance, does not always take values between −π and π. The plot of ∠H(ω) does always take values between −π and π. Notice that we can always find a multiple of 2π to add to the plot to obtain the formula given for ∠H(ω).
Example 6.5
Consider the RC circuit with input x given by the voltage source, and output y given by the voltage across the capacitor:
By Kirchoff’s voltage law
y(t) = x(t) − Ri(t)
where i is the current in the circuit. By the relationship between the current through, and
voltage across, a capacitor
i(t) = C dy (t). dt
So the relationship between the system input and output can be described by
y(t) + RC dy (t) = x(t). (6.5)
Since this is an LTI system, we know that if x(t) = ejωt then y(t) = H(ω)ejωt. We
substitute these quantities for x and y into (6.5) to try to find H(ω). Doing so we see that H(ω)ejωt + RC(jω)H(ω)ejωt = ejωt.
Solving for H(ω) gives
To find the phase response it is easier to first rewrite H(ω) as
H(ω)= 1 ·1−jωRC= 1−jωRC. 1+jωRC 1−jωRC 1+ω2R2C2
Then the phase response is
∠H(ω) = tan−1
H(ω) = 1 . 1+jωRC
−ωRC 1+ω2R2C2
1 1+ω2R2C2
= tan−1(−ωRC). 111
The magnitude response is
|H(ω)|= H(ω)H(ω)∗ = 1+jωRC1−jωRC = √1+ω2R2C2.
These functions are shown below for R = C = 1 and −4π ≤ ω ≤ 4π. |H(ω)|
ω −4π −2π 0 2π 4π
ω −4π −2π 0 2π 4π
6.2.1 Frequency response of systems defined by linear difference equations
Often discrete-time LTI systems are defined by linear difference equations subject to the con- dition of initial rest.
Frequency response of LTI system described by a difference equation
If the input x and output y of a DT LTI system are related by the linear constant-coefficient difference equation
a0y[n] + a1y[n − 1] + · · · + aN y[n − N] = b0x[n] + b1x[n − 1] + · · · + bM x[n − M] then the frequency response of the system is
b0 +b1e−jω +···+bMe−jMω H(ω)= a0 +a1e−jω +···+aNe−jNω.
To see why, we can put x[n] = ejωn for all integers n and y[n] = H(ω)ejωn for all integers n and substitute these into the difference equation. Doing so we obtain
a0H(ω)ejωn+a1H(ω)ejω(n−1)+···+aNH(ω)ejω(n−N) =b0ejωn+b1ejω(n−1)+···+bMejω(n−M).
Using the fact that ejω(n−m) = e−jωmejωn we can divide both sides by ejωn and solve for H(ω)
b0 +b1e−jω +···+bMe−jMω H(ω)= a0 +a1e−jω +···+aNe−jNω.
Example 6.6: Frequency response of DT finite impulse response LTI system
Inthespecialcasewherea0 =1anda1 =a2 =···=aN =0thenthesystemisafinite impulse response system described by
y[n] = b0x[n] + b1x[n − 1] + · · · + bM x[n − M].
Recall that in this case the impulse response is h[n] = b0δ[n]+b1δ[n−1]+···+bMδ[n−M]
only takes non-zero values for n = 0, 1, . . . , M . In this case the frequency response is H(ω)=b0 +b1e−jω +···+bMe−jMω.
It is a polynomial of degree M in e−jω.
6.2.2 Frequency response of systems defined by linear differential equations
The frequency response of continuous-time LTI systems defined by linear, constant-coefficient, differential equations (subject to the condition of initial rest) can be found via a similar approach used in the case of systems defined by difference equations.
Frequency response of LTI system described by a differential equation
If the input x and output y of a CT LTI system are related by the linear constant-coefficient differential equation
ay(t)+a dy+···+a dNy=bx(t)+b dx+···+b dMx 0 1 dt N dtN 0 1 dt M dtM
then the frequency response of the system is
b0 +b1(jω)+···+bM(jω)M H(ω) = a0 + a1(jω) + · · · + aN (jω)N .
To see why, we can put x(t) = ejωt for all t and y(t) = H(ω)ejωt for all t and substitute these into the differential equation. Using the fact that
dkx = (jω)kejωt and dky = H(ω)(jω)kejωt dxk dtk
a0H(ω)ejωt+a1H(ω)(jω)ejωt+···+aNH(ω)(jω)Nejωt =b0ejωt+b1(jω)ejωt+···+bM(jω)Mejωt. We can divide both sides by ejωt and solve for H(ω) to obtain
b0 +b1(jω)+···+bM(jω)M H(ω) = a0 + a1(jω) + · · · + aN (jω)N .
6.2.3 Frequency response of compositions of systems
Series interconnection
If two LTI systems are connected in series, then the frequency response is the product of the frequency responses of the individual systems. This is much simpler than the situation in the time-domain, where the impulse response of the series interconnection was the convolution of the impulse responses.
ejωt H1(ω)ejωt H1(ω)H2(ω)ejωt
Frequency response of series interconnection
If S1 and S2 are LTI systems with frequency responses H1 and H2 then the frequency
response of the series interconnection of S1 and S2 is H(ω) = H1(ω)H2(ω).
To see why, consider the CT case. If x1(t) = ejωt is the input of S1, then the output is y1(t) = H1(ω)ejωt. If x2(t) = ejωt is the input of S2, then the output is y2(t) = H2(ω)ejωt. By homogeneity of S2, if the input is y1(t) = H1(ω)x2(t), then the output will be H1(ω)y2(t) = H1(ω)H2(ω)ejωt. Therefore the frequency response is H1(ω)H2(ω).
Parallel interconnection
If two LTI systems are connected in parallel, then the frequency response is the sum of the frequency responses of the individual systems.
(H1(ω) + H2(ω))ejωt
Frequency response of parallel interconnection
If S1 and S2 are LTI systems with frequency responses H1 and H2 then the frequency response of the parallel interconnection of S1 and S2 is
H(ω) = H1(ω) + H2(ω).
To see why, simply note that if x(t) = ejωt is the input to the system, the output of the first system is y1(t) = H1(ω)ejωt and the output of the second system is y2(t) = H2(ω)ejωt. The overall output of the system is
y(t) = y1(t) + y2(t) = (H1(ω) + H2(ω))ejωt = H(ω)ejωt. Feedback interconnection
We now find the frequency response of the feedback interconnection of two LTI systems S1 and S2 shown below
Frequency response of feedback interconnection
Suppose S1 and S2 are LTI systems with frequency responses H1 and H2. If S1 and S2 are connected in feedback as shown above. Then the frequency response of the overall system
H(ω) = H1(ω) . 1 − H1(ω)H2(ω)
To see why this is true, we let x(t) = ejωt and y(t) = H(ω)ejωt. Then, because the frequency response of S2 is H2 we see that z(t) = H2(ω)H(ω)ejωt. Then
u(t) = x(t) + z(t) = ejωt + H2(ω)H(ω)ejωt = (1 + H2(ω)H(ω))ejωt. Finally since y is the output of S1 when the input is u we have that
This gives
Solving for H gives
y(t) = H1(ω)(1 + H2(ω)H(ω))ejωt = H(ω)ejωt.
H(ω) = H1(ω) + H1(ω)H2(ω)H(ω).
H(ω) = H1(ω) . 1 − H1(ω)H2(ω)
Summary of Section 6.2
We can explicitly find the frequency response of delay systems as well as LTI systems described by linear constant coefficient difference equations, and linear constant coefficient differential equations. Moreover, the frequency response of compositions of systems can be readily determined from the frequency responses of the component subsystems.
6.3 Summary
In this chapter we examined how LTI systems behave when the input is a unit complex expo- nential at a give frequency. We found that for CT LTI systems, if the input is ejωt then the output is H(ω)ejωt where H(ω) is a complex number called the frequency response at frequency ω. A similar result holds in discrete-time. The frequency response completely characterises
an LTI system, and is closely related to the impulse response of the system. We saw how to calculate the frequency response of LTI systems by explicitly examining the output when the input is a unit complex exponential at frequency ω.
Knowing how an LTI system transforms a single complex exponential, we could then under- stand how an LTI system transforms a periodic signal. This, and its implications, are the topic of the next Chapter.
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