Profs. Kamalabadi, Katselis, : 5 pm, March 6, 2022
UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering
ECE 310 Digital Signal Processing – Spring 2022
Rearranging to show H(z) = Y (z)/X(z) gives
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H(z)= Y(z) = 1 .
Homework 6 – Solution
1. A causal LSI system is described by the difference equation: y[n] − y[n − 1] = x[n].
(a) Determine the system’s transfer function H(z) (b) Determine the system’s unit pulse response h[n]
(c) Determine the system’s frequency response Hd(ω); is Hd(ω) = H(z)|z=ejω ? if not, explain why. Solution
(a): The transfer function can be found by taking the Z-transform of the system equation:
y[n]−y[n−1] = x[n]. n=−∞ n=−∞
By substituting in the definitions of Y (z) and X(z) and using the delay property, we find Y (z) − z−1Y (z) = X(z).
X(z) 1 − z−1
There are two possible ROC’s for H(z), given by |z| > 1 and |z| < 1. As the system is known to be
causal, we can deduce that the ROC is |z| > 1.
(b): The unit pulse response can be found via the transform table, which can be read off as
h[n] = u[n], using pair 2 of Table 10.
(c): The frequency response Hd(ω) is not equal to H(z = ejω). This can be readily seen by the ROC of H(z) being |z| > 1, barely not including the unit circle (defined as the set of all z such that |z| = 1). As such, H(z) does not apply at values of z satisfying |z| = 1, meaning the two will not be equal. This can be readily seen by the fact that
Hd(ω)= h[n]e−jωn= u[n]e−jωn=e−jωn̸=1−e−jω =H(z=ejω).
n=−∞ n=−∞ n=0 2. An LSI system is described by the difference equation
y[n] = x[n] + x[n − 10]
(a) Compute and sketch its magnitude and phase response (b) Determine its output to inputs
i. x[n] = cos π n+3sinπn+ π 10 310
ii. x[n]=10+5cos2πn+π 52
(a): The magnitude response can be found by taking the DTFT of the system equation to find Hd(ω) = Yd(ω)/Xd(ω):
Yd(ω)=Xd(ω)+e−j10ωXd(ω) =⇒ Hd(ω)=1+e−j10ω.
We want to find the magnitude and phase response, meaning we want to plot |Hd(ω)| and ∠Hd(ω). To find these, we try to write Hd(ω) = rejφ, where r = |Hd(ω)| ≥ 0 and φ = ∠Hd(ω) ∈ [−π, π]. This can be done using two methods.
This method will use the definitions of magnitude and phase to directly find r and φ. Starting with the
magnitude, and defining the rectangular form of Hd(ω) = a + jb, we know that |Hd(ω)| = and can find a and b directly, as
Hd(ω)=1+e−j10ω =1+cos(10ω)−jsin(10ω)=a+jb =⇒ |Hd(ω)| = (1 + cos(10ω))2 + sin2(10ω)
= 1 + 2 cos(10ω) + cos2(10ω) + sin2(10ω) = 2 + 2 cos(10ω) = 2| cos(5ω)|.
Additionally, we know that the phase of a complex number φ = tan−1(b/a), so ∠Hd(ω) = tan−1 sin(10ω) .
1 + cos(10ω)
This can be simplified further with some difficulty, but a more intuitive method can be used instead.
This method will manipulate Hd(ω) into its desired form. We begin by first extracting e−j5ω and using Euler’s identity:
Hd(ω) = 1 + e−j10ω = e−j5ω(ej5ω + e−j5ω) = 2 cos(5ω)e−j5ω.
This looks suspiciously like the desired magnitude-phase form, but the magnitude is not proper; cos(5ω) goes negative for certain values of ω on the interval [−π,π]. As such, corrections must be made to make this term a true magnitude. This can be done by invoking the following mathematical principle – a function f(ω) can be expressed in terms of its magnitude and sign as follows:
f(ω) = |f(ω)| × sign(f(ω)) = |f(ω)| ×
1 if f(ω) ≥ 0, −1 iff(ω)<0.
ej0 if cos(5ω) ≥ 0, Hd(ω) = 2| cos(5ω)|e−j5ω × ejπ if cos(5ω) < 0.
Figure 1: Magnitude and phase plot for 2(a).
As such, we can write
cos(5ω) = | cos(5ω)| × sign(cos(5ω)) = | cos(5ω)| ×
1 if cos(5ω) ≥ 0, −1 if cos(5ω) < 0.
This expression allows us to express cos(5ω) in terms of a proper magnitude and a unit-magnitude sign-carrying term that can be absorbed into the overall phase response. We can do this by writing
With this, we can plot the magnitude and phase readily, keeping in mind to add or subtract 2π to the phase to keep it within the range [−π, π] whenever necessary.
(b): The response of an LSI system to a sinusoidal input ejω0n is Hd(ω0)ejω0n. Using Euler’s identity, we can decompose cosine and sine into exponentials and use linearity to find the response to the two inputs.
i. We can re-express this input as
x[n]=1ejπ n+e−jπ n+ 3ej(πn+π )−e−j(πn+π ),
2 2j and write the corresponding output as
y[n]=1Hπejπn+1H−πe−jπn+ 3Hπej(πn+π)− 3H−πe−j(πn+π) d 10 d 10 d 3 10 d 3 10
2 10 2 10 2j 3 2j 3 We can evaluate Hd(ω) at ω = {± π ,±π} to find that
H π=H −π=0, and H π=1+e−j10π/3 =H∗−π. d10d10 d3 d3
10 10 310 310
Upon inserting into the above y[n], we find
y[n]=3 1+e−j10π/3ej(πn+π )− 3 1+ej10π/3e−j(πn+π )
2j 2j j(πn+ π ) −j(πn+ π )
5 2π j(2πn+π) where Hd(0) = 2, Hd(2π/5) = Hd∗(−2π/5) = 2, so
2π5 −j(2πn+π)
j(2π n+π ) −j(2π n+π ) y[n] = 20 + 5e 5 2 + 5e 5 2
3. The frequency response of an LSI system is
Hd(ω) = ωej sin ω,
|ω| ≤ π . Determine the system output y[n] for the following inputs:
(a) x[n]=3−10ej(π4n+45◦)+jn
(b) x[n]=3+10cos(π4n+45◦)−jn.
(a): Using the eigenfunction property, we know that
j(πn+ π −10π) −j(πn+ π −10π) =3e3 10−e3 10+e3 10 3 −e3 10 3
2j 2j π π π π 10π
=3sin 3n+10 +3sin 3n+10− 3 . ii. We can re-express this input as
2π π j0×n 5 j(2πn+π) 5 −j(2πn+π) x[n]=10+5cos n+ =10e +e5 2 +e 5 2
and as such can write
y[n] = 10Hd(0) + 2Hd 5 e 5 2
y[n]=3H (0)−10H πej(π4n+45◦) +H πjn dd4d2
where jn = ejπ/2n was used to show that the final term is an eigenfunction with frequency ω0 = π/2. √
We can use the given formula to show that Hd(0) = 0, Hd π = πej/ 2, and Hd π = πej, and 44 22
πj√1 j(πn+45◦) πjn y[n] = −10 e 2 e 4 + e j .
42 (b): Using the eigenfunction property, we know that
y[n]=3H (0)−5H πej(π4n+45◦) −5H −πe−j(π4n+45◦) −H πjn dd4d4d2
− 5 = 20 + 10 cos
We can use the given formula to show that Hd(0) = 0, Hd π = πej/ 2, Hd −π = −πe−j/ 2 and
Hd 2 =2e.Assuch,
π j√1 j(πn+45◦) y[n]=−5 e 2e 4
We can split x[n] into eigenfunctions as follows:
4 ej2n − The eigenfunction property can be applied to show that
1 − √1 e − j ω 4
442 5π j√1 j(πn+45◦) −j√1 −j(πn+45◦) π j n
=−e2e4 −e2e4 −ej 42
5ππ ◦ 1πjn =−4sin 4n+45+√2 −2ej.
4. The difference equation of a causal LSI system is given by
y[n]−√1 y[n−1]=x[n], −∞