You need to have Javascript enabled to view this page properly. If your browser does not support Javascript or you have chosen not to enable it, please return to the previous page and use the appropriate link to view non-script versions of this tutorial page.

6.02 Tutorial Problems: Frequency Domain & Filters

Problem .

Give an expression for the magnitude of a complex exponential with frequency φ, i.e., |e^jφ|. Hint: it's a numeric value independent of φ.

|e^jφ| = |cos(φ) + j*sin(φ)| = sqrt(cos²(φ) + sin²(φ)) = 1

Problem .

Prove the validity of the following formula, often referred to as the finite sum formala:
For α ≠ 1, do the following:
1. expand the sum
2. multiply expansion by (1–α)/(1–α)
3. simplify the numerator
Use the finite sum formula to show:

Simply apply the finite sum formula and use the fact that e^jπ = –1.

Problem .

Consider the following signals, which are periodic with period N. For each signal compute the spectral coefficients a_k. In 6.02, we usually choose N consecutive k's starting with -N/2.

x[n] = 1 + sin((2π/N)*n) + 3*cos((2π/N)*n) + cos(2*(2π/N)*n + π/2).
Rewrite the sin and cos componients using complex exponentials and gather e^jk(2π/N)n terms for the different k.
```
a_-2 = -(1/2)j
a_-1 = (3/2) - (1/2j) = (3/2) + (1/2)j
a₀ = 1
a₁ = (3/2) + (1/2j) = (3/2) - (1/2)j
a₂ = (1/2)j

a_k = 0  otherwise
```
Note the a_k are periodic with period N, so, for example, a_N = 1. For simplicity, we'll just list the spectal coefficients around 0.
x[n] = 5*cos(6πn + π) + 7*cos(3πn)
First determine the smallest N for which x[n] is periodic and then rewrite the frequencies in the form k(2π/N) for some k. If the resulting k is greater than N, you can replace it with k mod N.
In this example, the smallest N is 2:
```
x[n] = 5*cos(6(2π/2)n + π) + 7*cos(3(2π/2)n)  = 5cos(π) + 7*cos(1(2π/2)n)
```
Now we compute the two spectral coefficients (actually we can read them off the equation by inspection!):
```
a₀ = -5
a₁ = 7/2
```
x[n] is a square wave with period N=4 with the following values: x[0]=1, x[1]=1, x[2]=0, x[3]=0.
Compute the answer using the formula
```
a_k = (1/N)Σ_n=<N>x[n]e^-jk(2π/N)n
```
You can save yourself some work by remembering that a_k is the complex conjugate of a_-k when x[n] is real.
```
a_-2 = (1 + -1)/4 = 0
a_-1 = (1 + j)/4
a₀ = (1 + 1)/4 = 1/2
a₁ = (1 - j)/4
```
x[n] = cos(2πn/3)*sin(2πn/9)
The smallest N for which x[n] is periodic is 9. Expanding the product in terms of complex exponentials and gathering terms, we get
```
a_-4 = j/4
a_-2 = -j/4
a₀ = 0
a₂ = j/4
a₄ = -j/4

a_k = 0 otherwise
```
x[n] has exactly one non-zero value per period, i.e., x[m] ≠ 0 for some m and 0 otherwise. Compute the magnitude of a_k.
```
| a_k | = | (1/N)Σ_n=<N>x[n]e^-jk(2π/N)n |
       = (1/N)|x[m]| | e^-jk(2π/N)m |
       = |x[m]|/N
```
since the magnitude of e^jφ is 1 for any φ.

Problem .

If x[n] is real, even (i.e., x[n] = x[-n]) and periodic with period N, show that all the a_k are real.

a_k = (1/N)Σ_n=<N>x[n]e^-jk(2π/N)n

Noting that x[1] = x[-1], x[2] = x[-2], ..., if N is odd we can rewrite this as:

a_k = (1/N)[x[0] + x[1](e^jk(2π/N) + e^-jk(2π/N)) + x[2](e^jk(2π/N)2 + e^-jk(2π/N)2) + ...]
   = (1/N)[x[0] + x[1]*2cos(2πk/N) + x[2]*2cos(4πk/N) + ...]

which is a sum of real numbers. If N is even, there will be one negative index in the equation, n = -N/2, that won't be matched with a positive counterpart. But

x[-N/2]e^{-jk(2π/N)(-N/2)} = x[-N/2]e^jkπ = x[-N/2]*(-1)^k

which is also real.

Problem .

Suppose you're given the spectral coefficients a_k for a particular periodic sequence x[n]. Compute the spectral coefficients b_k for w[n] = x[n-α], i.e., x time shifted by α samples, in terms of the a_k.

b_k = (1/N)Σ_n=<N>w[n]e^-jk(2π/N)n
   = (1/N)Σ_n=<N>x[n-α]e^-jk(2π/N)n

This time, let's run the indicies of the summation from α to α+(N-1):

b_k = (1/N)Σ_{α ≤ n ≤ α+(N-1)}x[n-α]e^-jk(2π/N)n
   = (1/N)Σ_{0 ≤ n ≤ (N-1)}x[n]e^{-jk(2π/N)(n+α)}
   = (1/N)Σ_{0 ≤ n ≤ (N-1)}x[n]e^-jk(2π/N)ne^-jk(2π/N)α
   = e^-jk(2π/N)α(1/N)Σ_{0 ≤ n ≤ (N-1)}x[n]e^-jk(2π/N)n
   = a_ke^-jk(2π/N)α

Problem .

Consider an LTI system characterized by the unit-sample response h[n].

Give an expression for the frequency response of the system H(e^jΩ) in terms of h[n].
H(e^jΩ) = Σ_mh[m]e^-jΩm.
If h[0]=1, h[1]=0, h[2]=1, and h[n]=0 for all other n, what is H(e^jΩ)?
H(e^jΩ) = 1 + e^-2jΩ.
Let h[n] be defined as in part B and x[n] = cos(φn). Is there a value of φ such that y[n]=0 for all n and 0 ≤ φ ≤ π?
H(e^jΩ)=0 when φ = π/2.
Let h[n] be defined as in part B. Find the maximum magnitude of y[n] if x[n] = cos(πn/4).
|H(e^jπ/4)| = |1 + e^-jπ/2| = |1 - j| = sqrt(2).
Let h[n] be defined as in part B. Find the maximum magnitude of y[n] if x[n] = cos(-(π/2)n).
|H(e^jπ/2)| = |1 + e^-jπ| = |1 - 1| = 0.

Problem .

In answering the questions below, please consider the unit sample response and frequency response of two filters, H₁ and H₂, plotted below.

Note: the only nonzero values of unit sample response for H₁ are : h₁[0] = 1, h₁[1]=0, h₁[2]=1.

Note, the only nonzero values of unit sample response for H₂ are : h₂[0] = 1, h₂[1]=-sqrt(3), h₂[2]=1.

In answering the several parts of this review question consider four linear time-invariant systems, denoted A, B, C, and D, each characterized by the magnitude of its frequency response, |H_A(e^jΩ)|, |H_B(e^jΩ)|, |H_C(e^jΩ})|, and |H_D(e^jΩ)| respectively, as given in the plots below. This is a review problem, not an actual exam question, so similar concepts are tested multiple times to give you practice

Which frequency response (A, B, C or D) corresponds to a unit sample response given by
h[n] = α δ[n] - h₁[n]
and what is the numerical value of |α|?
Must be H_B as that is the only frequency response that has the same values at 0 and ±π and extremes at ±&pi/2.
|α|=2 as H₁(e^j0)=2 but H_B(e^j0)=0.
Which frequency response (A, B, C or D) corresponds to a unit sample response given by
h[n] = Σ_mh₁[m]h₂[n-m] for m = 0 to n
and what are the numerical values of h[2], h[3] and H(e^j0)?
Must be H_A since H₁(e^jΩ)H₂(e^jΩ) = H(e^jΩ), and therefore |H(e^jΩ)|=0 whenever H₁(e^jΩ)=0 or H₂(e^jΩ)=0.
H(e^j0) = H₁(e^j0)H₂(e^j0) = 2(2 - sqrt(3)) = 4 - 2sqrt(3)
h[n] = [1,0,1]*[1,-sqrt(3),1] so h[2] = 2 and h[3] = -sqrt(3).
Which frequency response (A, B, C or D) corresponds to a unit sample response given by
h[n] = α δ[n] - Σ_mh₁[m]h₂[n-m] for m = 0 to n
and what is the numerical value of |α|?
Since H_A is the frequency response for H₁*H₂, H_D must be the solution. It's the only frequency response with enough wiggles.
Since |H_A(e^jπ)| = |H₁(e^jπ)||H₂(e^jπ)|, |α| = 2(2 + sqrt(3)) = 4 + 2sqrt(3).
Which frequency response (A, B, C or D) corresponds to a unit sample response given by
h[n] = α δ[n] - h₂[n]
and what is the numerical value of |α|?
Must be H_C by elimination but also because none of the other frequency repsonse could be generated by a single magnitude shift of H₂(e^jΩ).
H_C(e^j0) = 0 so |α| = |H₂(e^j0)| = 2 - sqrt(3).
Suppose the input to each of the above four systems is
x[n]=0 for n < 0 and
x[n] = cos(nπ/6) + cos(nπ/2) + 1.0 for n ≥ 0
Which system (A, B, C or D) produced an output, y[n] below, and what is the value of y[n] for n > 10?

Must be H_A, as |H_A(e^jΩ)| = 0 for Ω=±π/6 and Ω=±π/2.
y[n] for n > 10 = |H_A(e^j0)|*1 = 4 - 2sqrt(3)

Suppose the input to each of the above four systems is

x[n]=0 for n < 0 and
x[n] = cos(nπ/6) + cos(nπ/2) + 1.0 for n ≥ 0

Which system (H₁ or H₂) produced an output, y[n] below, and what is the value of y[22]?

Must be H₁ since the H₂ system would eliminate cos(π/6), and since the output will eventually be a cosine offset by H₁(e^j0)*1.

First compute some useful H's from the given h[n]:

H(e^jΩ) = 1 + e^-j2Ω
H(e^j0) = 1 + 1 = 2
H(e^j(-π/6)) = 1 + e^jπ/3
H(e^j(π/6)) = 1 + e^-jπ/3
H(e^j(-π/2)) = 1 + e^jπ = 1 - 1 = 0
H(e^j(π/2)) = 1 + e^-jπ = 1 - 1 = 0

Now we can plug those into our equation for y[n] that uses the spectral coefficients for x[n] and the frequency response H:

y[n] = Σ_ka_kH(e^jk(2π/N))e^jk(2π/N)n
     = 2 + (1/2)H(e^j(-π/6))e^j(-π/6)n  + (1/2)H(e^j(π/6))e^j(π/6)n
     = 2 + (1/2)(1 + e^jπ/3)e^j(-π/6)n  + (1/2)(1 + e^-jπ/3)e^j(π/6)n
     = 2 + (1/2)e^j(-π/6)n + (1/2)e^{-j((π/6)n - π/3)} + (1/2)e^j(π/6)n + (1/2)e^{j((π/6)n - π/3)}
     = 2 + cos((π/6)n) + cos((π/6)n - π/3)

y[22] = 2 + cos(22π/6) + cos(22π/6 - π/3)
      = 2 + 0.5 - 0.5
      = 2

Problem .

Which frequency response (A, B, or C) corresponds to the following unit sample response, and what is max_Ω |H(e^jΩ)| for your selected filter? Please justify your selection.
Plan of attack: evaluate |H(e^jΩ)| at several frequencies and see what we find.
```
Ω=0: |H(e^j0)| = |Σh[n](1)| = 0
Ω=π: |H(e^jπ)| = |Σh[n](-1)ⁿ| = 6
```
The only plot with a frequency response of 0 at frequency 0 and something non-zero at frequency π is the one for filter A.
Looking at the frequency response for A, it has its largest magnitude at ±π, which we calculated above to be 6.
Which frequency response (A, B, or C) corresponds to the following unit sample response, and is max_Ω |H(e^jΩ)| > 6 for your selected filter? Please justify your answers.
Repeating the strategy of part A:
```
Ω=0: |H(e^j0)| = |Σh[n]| = 1 + .5 - 1.5 - 1.5 + .5 + 1 = 0
Ω=π: |H(e^jπ)| = |Σh[n](-1)ⁿ| = 1 -.5 - 1.5 + 1.5 +.5 - 1 = 0
```
The only plot with a frequency response of 0 at frequencies 0 and π is the one for filter B.
The maximum magnitude of the frequency response is given by:
```
|H(e^jΩ)| = |Σ_nh[n]e^-jΩn| ≤ Σ_n|h[n]|(1) ≤ 6.
```
So, no, it's not bigger than 6.
Suppose the input to each of the above three filters is x[n] = 0 for n < 0 and for n ≥ 0 is
x[n] = cos((π/3)n) + cos(πn) + 1.0
Which filter (A, B,or C) produced the output, y[n] below, and what is max_Ω |H(e^jΩ)| for your selected system?

The output reaches a steady state at y[n] = 6, with no changes in value after that point. So sinusoidal components of x[n] must have been filtered out, i.e., H(e^jπ/3) = 0 and H(e^jπ) = 0.
The only other component of x[n] is the constant 1, so that means that H(e^j0) = 6 since constant inputs have frequency 0.
The only plot that satisfies these conditions is C. From the plot of C's frequency response, the maximum magnitude occurs at frequency 0 and has the value 6 as we calculated above.
Six new filters were generated using the unit sample responses of filters A, B and C, denoted hA[n], hB[n], and hC[n] respectively. The unit sample responses of the new filters were gener- ated in the following way:
- h1[n] = hA[n] + hB[n]. Filters A and B in parallel.
- h2[n] = hA[n] + hC[n]. Filters A and C in parallel.
- h3[n] = hB[n] + hD[n]. Filters B and C in parallel.
- h4[n] = convolve(hA,hB). Filters A and B in series.
- h5[n] = convolve(hA,hC). Filters A and C in series.
- h6[n] = convolve(hB,hC). Filters B and C in series.
Which of the six new filters has the frequency response plotted below?

Only H_C in non-zero at frequency 0 and only H_A is non-zero at frequency π. So plot must correspond to new filter 2, which is filters A and C in parallel.

Which of the six new filters from the previous question has the frequency response plotted below?

Looking at the zeros in the plot above, the combination could have only come from two filters in series. Comparing the size of the response near frequency 0 with the response near frequency π, it must be filter B in series with filter C, i.e., new filter 6.