EEL3135: Homework #3 Solutions

Problem 1:(a) Compute the CTFT for the following signal: .

First, we use the trigonometric identity (easy to show by using the inverse Euler relations):

(S-1)

(S-2)

Therefore, the CTFT is given by:

(S-3)

(b) Compute for the following CTFT: . Your final answer should not include the imaginary number .

From the definition of the inverse Fourier transform:

(S-4)

(S-5)

(S-6)

(S-7)

(S-8)

(S-9)

x t( ) 2πt( )cos 3t( )cos 4πt( )cos+=

2πt( )cos 3t( )cos 12--- 3 2π–( )t[ ]cos 1

2--- 3 2π+( )t[ ]cos+=

2πt( )cos 3t( )cos 12--- 2π 3–( )t[ ]cos 1

2--- 3 2π+( )t[ ]cos+=

X f( ) 14--- δ f 2π 3–( )

2π--------------------+

δ f 2π 3–( )2π

--------------------– + 1

4--- δ f 3 2π+( )

2π--------------------+

δ f 3 2π+( )2π

--------------------– + + +=

12--- δ f 2+( ) δ f 2–( )+[ ]

x t( ) X f( ) 2ejπ 3⁄ δ f 4+( ) 2e jπ 3⁄– δ f 4–( )+=j

x t( ) X f( )ej2πft fd∞–

∞∫=

x t( ) 2ejπ 3⁄ δ f 4+( ) 2e jπ 3⁄– δ f 4–( )+[ ]ej2πft fd∞–

∞∫=

x t( ) 2ejπ 3⁄ e j8πt– 2e jπ 3⁄– ej8πt+=

x t( ) 2 e j 8πt π 3⁄–( )– ej 8πt π 3⁄–( )+[ ]=

x t( ) 4 e j 8πt π 3⁄–( )– ej 8πt π 3⁄–( )+2

---------------------------------------------------------------=

x t( ) 4 8πt π 3⁄–( )cos=

- 1 -

Problem 2:Consider the continuous-time signal:

, . (2-1)

(a) What is the fundamental frequency of this periodic signal?

Note that is already in the form of a Fourier series:

(S-10)

So, we do not need to compute the Fourier series coefficients through integration. Now, in the Fourier series, all the frequencies have to be integer multiples of . By inspection, Hz, so that the four terms in equation (2-1) correspond to , , and .

(b) Specify the Fourier series coefficients , , for .

By comparing equation (2-1) and (S-10), we can find by inspection:

(S-11)

(S-12)

[ ] (S-13)

(S-14)

Using the property that :

(S-15)

(S-16)

(S-17)

All other Fourier series coefficients are zero.

(c) Specify the Fourier series coefficients for the signal .

Starting with the complex Fourier series representation:

(S-18)

(S-19)

Therefore .

x t( ) 5 2πt( )cos 3 7πt( )sin 10 8πt π 4⁄–( )cos+ + += ∞– t ∞< <

f0

x t( )

x t( ) X0 2 Xk 2πkf0t Xk∠+( )cosk 1=

∞

∑+=

2πf0 f0 1 2⁄=k 0= k 2±= k 7±= k 8±=

Xk ∞– k ∞< < x t( )

Xk

X0 5=

X2 1 2⁄=

X7 3 2⁄ e jπ 2⁄– 3– j 2⁄= = 7πt( )sin 7πt π 2⁄–( )cos=

X8 10 2⁄ e j– π 4⁄ 5e j– π 4⁄= =

X k– Xk∗=

X 2– 1 2⁄=

X 7– 3j 2⁄=

X 8– 5ejπ 4⁄=

X'k x' t( ) 2x t 1+( )=

x t( ) Xkejπkt

k ∞–=

∞

∑=

2x t 1+( ) 2Xkejπk t 1+( )

k ∞–=

∞

∑ 2Xkejπk( )ejπkt

k ∞–=

∞

∑= =

Xk' 2Xkejπk=

- 2 -

Problem 3:Consider the periodic sawtooth waveform plotted below.

(a) Derive the complex Fourier series coefficients for this waveform.

Hints: Note that this is just a shifted version of the sawtooth wave in Figure 4 (page 5, Fourier Series notes); that is,

. (3-1)

Therefore, you do not have to derive the coefficients from scratch, although you are free to do so (not recommended though).

In the Fourier Series lecture notes, we derived the following Fourier series coefficients for the unshifted sawtooth waveform :

(S-20)

where the complex Fourier series representation is given by,

(S-21)

Substituting for in equation (S-21):

(S-22)

(S-23)

(S-24)

From equation (S-24), we observe that the Fourier coefficients for the shifted waveform are given by,

(S-25)

(S-26)

x1 t( )

-2 -1 0 1 2-1

-0.5

0

0.5

1

x1 t( )

t

t 1 4⁄–=

t 3 4⁄=

Xk

x t( )

x1 t( ) x t 1 4⁄–( )=

Xk

x t( )

Xkj 1–( )k( ) 2πk( )⁄

0

=k 0≠k 0=

x t( ) Xkej2πkt

k ∞–=

∞

∑=

t 1 4⁄–( ) t

x t 1 4⁄–( ) Xkej2πk t 1 4⁄–( )

k ∞–=

∞

∑=

x t 1 4⁄–( ) Xkej2πktej2πk 1 4⁄–( )

k ∞–=

∞

∑=

x t 1 4⁄–( ) Xke jπk 2⁄–( )ej2πkt

k ∞–=

∞

∑=

Xk'

Xk' Xke jπk 2⁄–=

Xk'j 1–( )ke jπk 2⁄–( ) 2πk( )⁄

0

=k 0≠k 0=

- 3 -

Equation (S-26) can be simplified to:

(S-27)

(The result in (S-27) can be verified by direct substitution of integers .)

(b) Plot the frequency spectrum for this waveform. That is, plot and as a function of , .

(c) Plot the first 5 terms of the trigonometric Fourier series for .

The trigonometric representation of the Fourier series for the shifted waveform is given by,

(S-28)

So the fifth-order approximation is given by,

(S-29)

(S-30)

(S-31)

Equation (S-31) is plotted below. (For MATLAB and Mathematica code, see the course web page.)

Xk'j k 1+( )( ) 2πk( )⁄

0

=k 0≠k 0=

k

Xk Xk∠ k 50– k 50≤ ≤

-40 -20 0 20 400

0.025

0.05

0.075

0.1

0.125

0.15

-40 -20 0 20 40

-1

0

1

2

3

Xk' Xk'

()

arg

k k

2– t 2≤ ≤

x t 1 4⁄–( ) X0' 2 Xk' 2πkt Xk'∠+( )cosk 1=

∞

∑+=

x t 1 4⁄–( ) 2 Xk' 2πkt Xk'∠+( )cosk 1=

5

∑≈

x t 1 4⁄–( ) 2πt π+( )cosπ

-------------------------------- 4πt π 2⁄–( )cos2π

-------------------------------------- 6πt( )cos3π

---------------------- 8πt π 2⁄+( )cos4π

--------------------------------------- 10πt π+( )cos5π

-----------------------------------+ + + +≈

x t 1 4⁄–( ) 2πt( )cos–π

------------------------- 4πt( )sin2π

--------------------- 6πt( )cos3π

---------------------- 8πt( )sin4π

---------------------– 10πt( )cos5π

-------------------------–+ +≈

-2 -1 0 1 2-1

-0.5

0

0.5

1

t

Truncated Fourier series

- 4 -

Problem 4:Below, there are two magnitude frequency spectra plotted, corresponding to the following two time-domain signals:

, (4-1)

, (4-2)

Specify and explain which signal corresponds to which magnitude frequency spectrum.

The equations below give the correct correspondences between the time and frequency domain plots:

(S-32)

(S-33)

Note that is given by,

. (S-34)

Also, note that the time limiting of signal causes the frequency content to spread out in the frequency domain.

x1 t( ) 4πt( )cos= ∞– t ∞< <

x2 t( ) 4πt( )cos u t( ) u t 5–( )–[ ]= ∞– t ∞< <

-15 -10 -5 0 5 10 15 -15 -10 -5 0 5 10 15f

Xb f( )Xa f( )

f

x1 t( ) Xa f( )⇔

x2 t( ) Xb f( )⇔

Xa f( )

Xa f( ) 12---δ f 2+( ) 1

2---δ f 2–( )+=

x2 t( )

- 5 -

Problem 5:Below, the magnitudes of the DTFT are plotted for two sampled signals and given by,

where , and (5-1)

where . (5-2)

(a) Are the DTFTs plotted as a function of the frequency variable or frequency (in Hz)?

Since we should expect peaks in the DTFTs at Hz, we note that the DTFTs are plotted as a function of frequency .

(b) Specify and explain which discrete-time signal corresponds to which magnitude frequency spectrum.

The equations below give the correct correspondences between the time and frequency domain plots:

(S-35)

(S-36)

The signal results from longer sampling than and therefore its frequency content is more localized about the dominant frequencies of Hz.

x1 n[ ] x2 n[ ]

x1 n[ ]xc n 10⁄( )

0

=n 0 1 … 98 99, , , ,{ }∈elsewhere

xc t( ) 4πt( )cos=

x2 n[ ]xc n 10⁄( )

0

=n 0 1 … 48 49, , , ,{ }∈elsewhere

xc t( ) 4πt( )cos=

θ f

2±f

-4 -2 0 2 40

10

20

30

40

50

-4 -2 0 2 40

5

10

15

20

25

Magnitude spectrum A (DTFT) Magnitude spectrum B (DTFT)

x1 t( ) spectrum A ⇔

x2 t( ) spectrum B ⇔

x1 n[ ] x2 n[ ]2±

- 6 -

Problem 6:Below, the magnitude of the DTFT is plotted for the sampled signal given by,

where . (6-1)

(a) Label the horizontal axis where indicated, assuming the DTFT is plotted as a function of .

(b) Label the horizontal axis where indicated, assuming the DTFT is plotted as a function of (in Hz).

Note the correct labeling of the horizontal axis as a function of both and . Remember that the DTFT is periodic with period , and that, as a function of frequency, the spectrum of the continuous-time waveform is replicated at , where is an integer.

x n[ ]

x n[ ]xc n 20⁄( )

0

=n 0 1 … 98 99, , , ,{ }∈elsewhere

xc t( ) 4πt( )cos=

θ

f

Magnitude spectrum (DTFT)

0

10

20

30

40

50

θ

f

4π– 2π– 0 2π 4π

40Hz20Hz0Hz20– Hz40– Hz

θ f2π

kfs± k

- 7 -

Problem 7:Below, two magnitude frequency spectra are plotted for two signals and given by,

, and, (7-1)

(7-2)

(a) To what variable does the horizontal axis in each plot correspond?

The magnitude spectra are plotted as a function of frequency in Hertz.

(b) Specify and explain which signal corresponds to which magnitude frequency spectrum.

The equations below give the correct correspondences between the time and frequency domain plots:

(S-37)

(S-38)

Note that the time-limited continuous-time signal has dominant frequencies at , while the sampled version of the signal has the original spectrum replicated at , where is an integer.

(c) Specify to which frequency transform each magnitude spectrum corresponds (e.g. CTFT, DTFT, DFT).

Magnitude spectrum A corresponds to the DTFT, while magnitude spectrum B corresponds to the CTFT.

x1 t( ) x2 n[ ]

x1 t( ) 4πt( )cos u t( ) u t 5–( )–[ ]=

x2 n[ ]x1 n 10⁄( )

0

=n 0 1 … 48 49, , , ,{ }∈elsewhere

f

x1 t( ) spectrum B ⇔

x2 n[ ] spectrum A ⇔

x1 t( ) 2Hz±kfs± k

-20 -10 0 10 20

-20 -10 0 10 20

Magnitude spectrum A

Magnitude spectrum B

- 8 -

Problem 8:Below, the magnitudes of the DTFT are plotted for two sampled signals and given by,

where , and (8-1)

where . (8-2)

(a) Specify and explain which discrete-time signal corresponds to which magnitude frequency spectrum.

(b) Indicate to which frequency (in Hz) each of the dominant peaks in the plots corresponds.

The equations below give the correct correspondences between the time and frequency domain plots:

(S-39)

(S-40)

Note that the sampling frequency of 3Hz for is below the Nyquist sampling frequency of 4Hz, and so aliasing occurs. The peaks for spectrum A correspond to , where is an integer. On the other had, the sampling frequency of 10Hz for is above the Nyquist sampling frequency of 4Hz, and so aliasing does not occur. The peaks for spectrum B correspond to:

(in Hertz) (S-41)

where is an integer.

To understand these plots, you can perform a similar analysis as for problems 1 and 2 on homework #2, assuming infinite-length sampling of .

x1 n[ ] x2 n[ ]

x1 n[ ]xc n 3⁄( )

0

=n 0 1 … 28 29, , , ,{ }∈elsewhere

xc t( ) 1 2 4πt( )cos+=

x2 n[ ]xc n 10⁄( )

0

=n 0 1 … 28 29, , , ,{ }∈elsewhere

xc t( ) 1 2 4πt( )cos+=

0

0

Magnitude spectrum A

Magnitude spectrum B

x1 n[ ] spectrum A ⇔

x2 n[ ] spectrum B ⇔

x1 n[ ]kHz± k

x2 n[ ]

2– 10k+ 0 10k+ 2 10k+, ,{ }

k

xc t( )

- 9 -

Problem 9:On the next page, the shape of ten magnitude frequency spectra are plotted as a function of frequency (in Hertz) corresponding to ten of the following eleven time-domain signals:

, (9-1)

, (9-2)

, (9-3)

, (9-4)

, (9-5)

, (9-6)

, (9-7)

, (9-8)

, (9-9)

, (9-10)

, (9-11)

(Note that the first three signals are continuous-time signals, while the last eight signals are discrete-time signals, and that one of the above signals does not correspond to any of the magnitude spectra shown.)

(a) For each of the above time-domain signals, assign its corresponding frequency spectrum (i.e. A, B, C, D, E, F, G, H, I, J), or indicate “none” if none of the frequency spectra correspond to a particular signal.

See the above labeling. The reason the DTFT of signal does not have multiple peaks is because of the discrete sequence generated by a sampling frequency of Hz is given by:

, . (S-42)

That is, it represents a discrete time pulse of width 24. Also, note the aliasing for signals and .

(b) For each frequency spectrum, label it as either a CTFT or a DTFT.

See the above labeling.

f

B/CTFT x1 t( ) 1 6πt( )sin+= ∞– t ∞< <

H/CTFT x2 t( ) x1 t( ) u t( ) u t 2–( )–[ ]= ∞– t ∞< <

E/CTFT x3 t( ) x1 t( ) u t( ) u t 3–( )–[ ]= ∞– t ∞< <

C/DTFT x4 n[ ] x1 n 10⁄( )= ∞– n ∞< <

D/DTFT x5 n[ ] x1 n 10⁄( ) u n[ ] u n 20–[ ]–( )= ∞– n ∞< <

J/DTFT x6 n[ ] x1 n 20⁄( ) u n[ ] u n 20–[ ]–( )= ∞– n ∞< <

F/DTFT x7 n[ ] x1 n 10⁄( ) u n[ ] u n 40–[ ]–( )= ∞– n ∞< <

A/DTFT x8 n[ ] x1 n 6⁄( ) u n[ ] u n 24–[ ]–( )= ∞– n ∞< <

G/DTFT x9 n[ ] x1 n 5⁄( ) u n[ ] u n 20–[ ]–( )= ∞– n ∞< <

I/DTFT x10 n[ ] x1 n 4⁄( ) u n[ ] u n 16–[ ]–( )= ∞– n ∞< <

none x11 n[ ] x1 n 3⁄( ) u n[ ] u n 12–[ ]–( )= ∞– n ∞< <

x8 n[ ]fs 6=

x8 n[ ] 1 6πn6

---------- sin+ 1 πn( )sin+ 1= = = n 0 1 … 22 23, , , ,{ }∈

x9 n[ ] x10 n[ ]

- 10 -

-20 -10 0 10 20 -20 -10 0 10 20

-20 -10 0 10 20 -20 -10 0 10 20

-20 -10 0 10 20 -20 -10 0 10 20

-20 -10 0 10 20 -20 -10 0 10 20

-20 -10 0 10 20 -20 -10 0 10 20

Magnitude frequency spectra (as function of frequency f)

A B

C D

E F

G H

I J

Problem 9 figures

- 11 -

Problem 10:(a) Compute the DTFT of the following discrete-time signal:

(10-1)

From the definition of the DTFT:

(S-43)

(S-44)

(b) For , sketch , .

For ,

(S-45)

See the magnitude plot below.

(c) For , sketch , .

See the phase plot above.

(d) Compute the DTFT of the following discrete-time signal:

(10-2)

and plot .

From the definition of the DTFT:

(S-46)

X ejθ( )

x n[ ] δ n n0–[ ]=

X ejθ( ) x n[ ]e jnθ–

n ∞–=

∞

∑=

X ejθ( ) δ n n0–[ ]e jn0θ– e jn0θ–= =

n0 2= X ejθ( ) π– θ π< <

n0 2=

X ejθ( ) e j2θ–=

-3 -2 -1 0 1 2 30

0.5

1

1.5

2

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

θ

X ejθ( ) X ejθ( )∠

θ

n0 2= X ejθ( )∠ π– θ π< <

X ejθ( )

x n[ ] δ n 3–[ ] δ n 3+[ ]+=

X ejθ( )

X ejθ( ) δ n 3–[ ]e j3θ– δ n 3+[ ]e j3θ;+ e j3θ– e j3θ;+ 2 3θ( )cos= = =

- 12 -

Problem 11:(a) Assume the magnitude plot of the DFT of a real-valued, discrete-time signal is given by the

plot below, and that was sampled from a continuous-time signal at a sampling frequency of 10Hz. Specify the set of discrete-time signals consistent with .

The length of the DFT is , and the sampling frequency is given as Hz. Therefore,

Hz (S-47)

Hz (S-48)

Hz (S-49)

Hz (S-50)

Since we are not given the phase of the DFT or the absolute magnitude, the set of discrete-time signals that are consistent with is given by:

(S-51)

where , and can be any real-valued scalars.

(b) Suppose that and . Plot for .

For all the zero DFT components . For the other two nonzero components:

(S-52)

(S-53)

See the plot below for the phase DFT as a function of index .

X k( ) x n[ ]x n[ ]

X k( )

0 2 4 6 8 10 12 14 16 18

k

X k( )

N 20= fs 10=

k 4= ⇔ f 4 10 20⁄( ) 2= =

k 16= ⇔ f 16 20–( ) 10 20⁄( ) 2–= =

k 8= ⇔ f 8 10 20⁄( ) 4= =

k 12= ⇔ f 12 20–( ) 10 20⁄( ) 4–= =

x n[ ]X k( )

x n[ ] A 4πn10

---------- α+ cos 2 8πn

10---------- β+

cos+ A 2πn5

---------- α+ cos 2 4πn

5---------- β+

cos+= =

A α β

X 16( )∠ π 4⁄= X 8( )∠ π 3⁄–= X k( )∠ k 0 1 … 18 19, , , ,{ }∈

X k( )∠ 0=

X 4( )∠ X 16( )∠– π 4⁄–= =

X 12( )∠ X 8( )∠– π 3⁄= =

k

0 2 4 6 8 10 12 14 16 18-3

-2

-1

0

1

2

3

k

X k( )∠

- 13 -

(c) Sketch the magnitude DFT as a function of frequency (in Hertz).

See the plot below for the magnitude DFT as a function of frequency .

(d) Which of the following frequencies cannot be represented exactly by this DFT?

Hz, Hz, Hz. (11-1)

The frequency resolution of this DFT is Hz. Therefore, Hz cannot be represented by this DFT. Also, since the sampling frequency is 10Hz, the maximum frequency that can be represented by this DFT is Hz. Therefore, Hz cannot be represented by this DFT.

f

-4 -3 -2 -1 0 1 2 3 4

f

magnitude spectrum

f 0.5= f 1.75= f 6=

10 20⁄( ) 0.5= f 1.75=

10 2⁄( ) 5= f 6=

- 14 -

Problem 12:Below, the magnitudes of the DFT are plotted for two sampled signals and given by,

where , and (12-1)

where . (12-2)

(a) Specify and explain which discrete-time signal corresponds to which magnitude frequency spectrum (DFT).

The equations below give the correct correspondences between the time and frequency domain plots:

(S-54)

(S-55)

This is because the sampled signal has frequency 21/20Hz, which is an integer multiple of Hz, while the sampled signal has frequency 1Hz, which is not an integer multiple of

Hz.

(b) For each of the indexes below, indicate the corresponding frequency :

, , . (12-3)

For the DFT, the frequency resolution is given by , which for this example is given by Hz. Then,

(S-56)

(S-57)

(S-58)

(c) For each of the frequencies below, indicate whether or not the DFT can represent those frequencies exactly:

Hz, Hz, Hz, Hz. (12-4)

The DFT can only represent integer multiples of Hz exactly. Therefore, 0.35Hz and 1.4Hz can be represented exactly in the DFT representation.

x1 n[ ] x2 n[ ]

x1 n[ ]xc1 n 7⁄( )

0

=n 0 1 … 18 19, , , ,{ }∈elsewhere

xc1 t( ) 2πt( )cos=

x2 n[ ]xc2 n 7⁄( )

0

=n 0 1 … 18 19, , , ,{ }∈elsewhere

xc2 t( ) 2π 21 20⁄( )t( )cos=

x1 n[ ] Xa k( )⇔

x2 n[ ] Xb k( ) ⇔

xc2 t( )fs N⁄ 7 20⁄= xc1 t( )7 20⁄

k f

k 0= k 3= k 17=

fs N⁄ 7 20⁄

k 0={ } 0Hz{ }⇔

k 3={ } 3 7 20⁄( )Hz 21 20⁄ Hz={ }⇔

k 17={ } 3 17 20–( ) 20⁄ Hz 21 20⁄ Hz–={ }⇔

f

f 0.35= f 1= f 1.4= f 1.5=

7 20⁄ 0.35=

0 5 10 15 20 0 5 10 15 20

Xa k( )

k k

Xb k( )

- 15 -

Problem 13:

(a) Match each discrete-time signal to its corresponding magnitude DTFT representation. Explain your answer.

, , (S-59)

Explanation: is a windowed pulse function, and should therefore exhibit less spectral leakage than the other two signals; is spread out more in the time domain than , and therefore will result in a more focused representation in the frequency domain.

(b) Match each of the following discrete-time signals,

, (13-1)

, (13-2)

, (13-3)

, (13-4)

, (13-5)

, (13-6)

x1 n[ ] Xc ejθ( )⇔ x2 n[ ] Xa ejθ( )⇔ x3 n[ ] Xb ejθ( )⇔

x1 n[ ]x2 n[ ] x3 n[ ]

-20 -10 0 10 20 -3 -2 -1 0 1 2 3

-20 -10 0 10 20 -3 -2 -1 0 1 2 3

-20 -10 0 10 20 -3 -2 -1 0 1 2 3

x1 n[ ]

x2 n[ ]

n θ

θ

Xa ejθ( )

Xb ejθ( )

x3 n[ ]

n

n θ

Xc ejθ( )

x1 n[ ] x n 3⁄( ) u n[ ] u n 15–[ ]–( )= ∞– n ∞< <

x2 n[ ] x n 3⁄( ) u n[ ] u n 30–[ ]–( )= ∞– n ∞< <

x3 n[ ] x n 8⁄( ) u n[ ] u n 40–[ ]–( )= ∞– n ∞< <

x4 n[ ] x n 8⁄( ) u n[ ] u n 80–[ ]–( )= ∞– n ∞< <

x5 n[ ] x n 20⁄( ) u n[ ] u n 100–[ ]–( )= ∞– n ∞< <

x6 n[ ] x n 20⁄( ) u n[ ] u n 200–[ ]–( )= ∞– n ∞< <

- 16 -

where , , to its corresponding magnitude DTFT representation below. Recall that . Explain your answer.

, , , , ,

(S-60)

Explanation: Signals and are undersampled and should exhibit peaks at Hz and 0Hz [easily derivable through idealized aliasing analysis], corresponding to values,

(S-61)

Signals and are sufficiently sampled and should exhibit peaks at Hz and 0Hz, corresponding to values,

(S-62)

Similarly, signals and are sufficiently sampled and should exhibit peaks at,

(S-63)

The above analysis helps us narrow down the potential matches for each time-domain signal to two. Exact matches between time-domain signals and frequency domain representations are made by realizing that longer-sampled sequences in the time-domain will result in more focused representations in the frequency domain.

x t( ) 1 4πt( )cos+= ∞– t ∞< <θ 2πf( ) fs⁄=

-3 -2 -1 0 1 2 30

20

40

60

80

-3 -2 -1 0 1 2 30

5

10

15

20

25

30

-3 -2 -1 0 1 2 30

50

100

150

200

-3 -2 -1 0 1 2 30

10

20

30

40

-3 -2 -1 0 1 2 30

2.5

5

7.5

10

12.5

15

-3 -2 -1 0 1 2 30

20

40

60

80

100

θ

Xb ejθ( )

θ

Xd ejθ( )

θ

Xf ejθ( )

θ

Xe ejθ( )

θ

Xc ejθ( )

θ

Xa ejθ( )

x1 n[ ] Xa ejθ( )⇔ x2 n[ ] Xf ejθ( )⇔ x3 n[ ] Xd ejθ( )⇔ x4 n[ ] Xe ejθ( )⇔ x5 n[ ] Xb ejθ( )⇔

x6 n[ ] Xc ejθ( )⇔

x1 n[ ] x2 n[ ] 1±θ

θ 2πf( ) fs⁄ 2π 3⁄± 0,= =

x3 n[ ] x4 n[ ] 2±θ

θ 2πf( ) fs⁄ 4π 8⁄± 0, π 2⁄± 0,= = =

x5 n[ ] x6 n[ ]

θ 2πf( ) fs⁄ 4π 20⁄± 0, π 5⁄± 0,= = =

- 17 -

Problem 14:(a) Write a computer program (MATLAB, Mathematica, etc.) to plot the magnitude and phase (as a function of

) of the DTFT for the following discrete-time signal:

where . (14-1)

(a) Now plot the magnitude and phase (as a function of ) of the DTFT for the following discrete-time signal:

(14-2)

where is given in (14-1) above.

See the plots below. Note that the shift in equation (14-2) only affects the phase part of the DTFT, not the magnitude part of the DTFT. (For MATLAB and Mathematica code, see the course web page.)

θ

x1 n[ ]xc n 10⁄( )

0

=n 0 1 … 20, , ,{ }∈elsewhere

xc t( ) 2πt( )cos 3 4πt( )cos+=

θ

x2 n[ ] x1 n 10+[ ]=

x1 n[ ]

-3 -2 -1 0 1 2 30

5

10

15

20

25

30

35

-3 -2 -1 0 1 2 30

0.5

1

1.5

2

2.5

3

-3 -2 -1 0 1 2 30

5

10

15

20

25

30

35

-3 -2 -1 0 1 2 3

-2

-1

0

1

2X1 ejθ( )

X2 ejθ( )

X1 ejθ( )∠

X2 ejθ( )∠

θ θ

θ θ

(a)

(b)

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