Tutorial Solutions PDF
| Title | Tutorial Solutions |
|---|---|
| Author | sonia Mani |
| Course | Signals And Systems |
| Institution | University of Western Australia |
| Pages | 74 |
| File Size | 3.6 MB |
| File Type | |
| Total Downloads | 15 |
| Total Views | 142 |
Summary
Download Tutorial Solutions PDF
Description
ENSC3015
Signals and Systems
Tutorial Solutions Contents Tutorial 0 (Solutions) Pre-requisites .................................................................................................................. 2 Tutorial 1 (Solutions) Introduction to Signals and Systems .............................................................................. 6 Tutorial 2 (Solutions) Continuous-Time Systems ........................................................................................... 12 Tutorial 3 (Solutions) Discrete-Time Systems ................................................................................................ 19 Tutorial 4 (Solutions) Laplace Transform ....................................................................................................... 23 Tutorial 5 (Solutions) z-Transform .................................................................................................................. 30 Tutorial 6 (Solutions) Introduction to Fourier Analysis .................................................................................. 41 Tutorial 7 (Solutions) Fourier Series ............................................................................................................... 42 Tutorial 8 (Solutions) Fourier Transform ........................................................................................................ 49 Tutorial 9 (Solutions) Sampling....................................................................................................................... 59 Tutorial 10 (Solutions) DTFS and DTFT ........................................................................................................ 67
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 1
ENSC3015
Signals and Systems
Tutorial 0 (Solutions) Pre-requisites Express the following numbers in polar complex form: (i) 1 + (ii) −4 + 3 (iii) + 1 (iv) (1 + )(−4 + 3) Answers: (i) = 1 + ≡ + , and = where = √2 + 2 = √12 + 12 = √2
1.
= tan−1 �� = tan−1 � 1� = , so that 1 + = √2 /4 = 1.414∠45 4
(ii)
1
= −4 + 3, so = �(−4)2 + 32 = √25 = 5 and 3 = atan2(3, −4) = − tan−1 � � = − 0.643 = 2.498, thus −4 + 3 = 5 2.498 4
(iii)
(iv)
= + 1 = cos(1) + sin(1) + 1, so = �(cos(1) + 1)2 + sin2 (1) = 1.755 and = tan−1 �
sin(1) � cos(1) +1
= 0.5, so + 1 = 1.755 /2
Using the results from (i) and (ii) we have ⁄ 2.498 4 (1 + )(−4 + 3) = �√2 ��5 � = 7.7071 3.283
Express the following numbers in Cartesian form: (i) (1 + )(−4 + 3) (ii) /4 + 2 −/4 Answers: (1 + )(−4 + 3) = (−4 − 3) + (−4 + 3) = −7 − (i)
2.
(ii)
/4 = cos � � + sin �4 � = 4
Thus:
/4
+ 2
/4
1
+ , and −/4 = cos � � − sin �4 � =
√2 1+2−2
√2
= +
√2
√2 3
=
1
√2
−
1
√2
4
Given 1 = 3 + 4 and 2 = 2 /4 : (i) Determine |1 |2 (ii) Calculate 1 + 2 , express your answer in Cartesian form. (ii) Calculate 1 2 , express your answer in Cartesian form. Answers:
1
√2
−
1
√2
3.
(i) Convert 1 to polar form: 1 = 1 1 , where 1 = √25 = 5, and 1 = tan−1 � =
0.927, thus 1 = 5 0.927, then:
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
3
4
Page 2
ENSC3015
Signals and Systems |1 |2 = 2 = 25
(ii) Convert 2 to Cartesian form: 2 = 2 cos � � + 2 sin �� = 1.414 + 1.414. Thus: 1 + 2 = (3 + 4) + (1.414 + 1.414) =4 4.414 + 5.4414
⁄ 4 1 2 = �5 0.927 ��2 � = 10 1.713 Converting to Cartesian form: 1 2 = 10 cos(1.713) + 10 sin(1.713) = −1.414 + 9.899
(iii) Use polar form:
Using the complex plane: (i) Express −1 to provide the most general expression in polar complex form (ii) Evaluate and plot the distinct solutions for ()4 = −1. (iii) If = √−1, what is � Answers: (i) −1 = where = 1 and = atan2(0, −1) = so that −1 = , but the most general expression involves any and all possible (rather than over the principal range of [0,2) ), in which case: −1 = (+2) , for any integer 4.
(ii) Since ()4 = (+2) then = � (+2) � = (/4+/2) which results in 4 unique positions on the complex plane (over the range = [0,2) ) for = 0,1,2,3, that is: = (/4) = (3/4) = (5/4) = (7/4) 1/4
2 +2) 2 then � (⁄ � = ⁄ 1 = (iii) Since = /2 ≡ ⁄( unique for = 0,1 or
23/08/2019
� = (
4
1⁄2 2 +2)
�
⁄ +) 4 = ( which is
Dr. Roberto Togneri, E&E Eng, UWA ⁄ )
⁄ ) 4 ⁄ +) 4 → ±� = . ( = (
Page 3
1
1 �
ENSC3015
Signals and Systems
Determine an expression for an exponentially decaying sinusoid that oscillates three times per second and whose amplitude envelope decreases by 50% every 2 seconds. Answers: The general form is () = − cos(). At = 0, − = 1 which is 0.5 at = 2, that 0.5 = −2 → = 0.5 ln(2) = 0.3466. If the sinusoid oscillates 3 times a second then = 3 Hz and = 2 = 6, hence: () = −0.3466 cos(6) 5.
You are working on a digital communications receiver which requires a pair of quadrature signals, cosn(Ω) and sin(Ω) to be generated at each integer sample index . One approach is to first generate the signal = Ω and then recursively multiply the signal by itself to generate 2 , 3 , …. Show how this method generates the pair of quadrature signals. Answers: 6.
2 = � Ω� = 2Ω 2
= Ω = cos(Ω) + sin(Ω)
And in general:
cos(Ω) = Re( ) sin(Ω) = Im( )
So that:
Consider the discrete-time sinusoid cos(Ω) where is the integer sample index. Show that if Ω = 2⁄ , for integer the sinusoidal signal is periodic with period and there is one sinusoidal cycle per . Now consider Ω = 2/ for integer, where and have no factors in common, and analyse the periodicity. Answers: 7.
Let Ω = 2/, then:
cos(Ω) = cos �
2 �
Consider 1 = + then: 21 2 2 2( + ) � cos � � = cos � + 2� = cos � � = cos � And for every samples the sinusoid exhibits one cycle of 2 and repeats the same sequence (periodic). Let Ω = 2/, then:
Consider = + then: 23/08/2019
cos(Ω) = cos �
2 �
Dr. Roberto Togneri, E&E Eng, UWA
Page 4
ENSC3015
Signals and Systems
2 2 2� 2( + ) cos � Since and have no factors �in=common is = so = cos � for periodicity cos � then the only�solution + that: 2 2� 2 cos � + 2� = cos � � = cos � And for every samples the exhibits cycles of 2 and repeats the same sequence sinusoid (periodic).
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 5
ENSC3015
Signals and Systems
Tutorial 1 (Solutions) Introduction to Signals and Systems Determine whether the following signals are periodic. If fundamental period. (a) () = (cos(2))2 +1 ∞ (b) () = ∑ ( − 3) − +1 where () = � =−∞ 0 (c) [] = (−1) 2 (d) [] = (−1) (e) [] = cos(2) (f) [] = cos(2) Answers: 1.
they are periodic, find the −1 ≤ ≤ 0 0 ≤ ≤ 1 othewise
(a) Since cos(2) where = 2 has a period of = = 1 second. With (cos(2))2 the symmetric negative swings become positive effectively doubling the cycles = 2, making it periodic with 0 = 0.5 seconds (this is also evident from the identity (cos(2))2 = 0.5 cos 4 + 0.5). 2
(b) The signal () = ⋯ + ( + 6) + ( + 3) + () + ( − 3) + ( − 6) + ⋯ and is periodic with 0 = 3 seconds. (c) The sequence 0 = 2 samples.
[] = (−1)
is
the
pattern
−1,1, −1,1, −1,1
which
is
periodic
with
1 , −1,1, −1,1, … } which is-periodic with 0 = 2 samples. To see this (d) The sequence [] = (−1) is { ⏟ 2
=0
we need to show 2 produces alternating even/odd integer sequences: e.g. ()2 = , ()2 = .
(e) [] = cos(2) where Ω = 2 ≠ 2 for any integer value {, 0 }, hence non-periodic 0
for (f) [] = cos(2) where Ω = 2 = 2 = 1, 0 = 1 hence it is periodic with 0 = 1 sample. 0
Calculate: (a) The average power of () = cos( + ) 0.5[cos() + 1] (b) The total energy of () = � 0 HINT: Use the identity: cos 2 = 2 cos 2 − 1 Answers:
2.
(a)
−/ ≤ ≤ / otherwise
1 /2 2 /2 � cos 2 ( + ) = lim � |()|2 = = lim →∞ →∞ −/2 −/2
2 ⁄2 2 ⁄2 2 /2 � 1 � cos(2 + 2) � {cos(2 + 2) + 1} + lim = lim →∞ 2 →∞ 2 →∞ 2 −⁄ 2 −/2 −⁄ 2 2 2 = +0= 2 2
= lim
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 6
ENSC3015 (b)
3.
Signals and Systems
⁄ ∞ 1 ⁄ = � |()|2 = 1 −⁄ (cos() + 1)2 � (cos 2 () + 2 cos() + 1) = � ⁄ 4 2 1−∞ 0 = � ({0.5cos(2) + 0.5} + 2 cos() + 1) 2 0 ⁄ 1 ⁄ 1 1 ⁄ 3 1 = � cos(2) + � 2 cos() + � 2 2 0 2 0 2 2 0 3 1 3 = 0 + 0 + � �� �� � = 4 2 2
(a) Sketch () = (10 − 5) where:
(b) Sketch ()( − 1) where: (c) Sketch the waveform () = ( + 1) − 2() + ( − 1) Answers: (a) (10 − 5), shift first then scale to yield:
(b)
(c) 23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 7
ENSC3015
4.
Signals and Systems
Let [] and [] be given as below:
Carefully sketch: (a) [2], (b) [3 − 1], Answers:
(c) [ − 2] + [ + 2]
(a) [2] ≡ [] keep every 2nd sample, i.e. = −2, −1,0,1,2 maps to = −4, −2,0,2,4 [] = 0,2,0,2,0
(b) [3 − 1] = [] so = −1,0,1,2 maps to = −4, −1,2,5 [] = 0,1,2,0
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 8
ENSC3015
Signals and Systems
(c) [] delayed by 2 samples is added to [] advanced by 2 samples:
5.
(a) Consider the sinusoidal signal [] = sin(0.6). What is the fundamental period 0 and how many sinusoidal periods does this require? (b) Show that the two LTI systems below are equivalent. Which configuration is simpler to implement?
Answers:
3 = 0.3 = Ω = 0.6 ≡ 2 → 10 0 0 hence 0 = 10. This means a period is reached when sin(0.60 ) = sin(6) which implies 3 sinusoidal periods.
(a) We have that:
(b) For the left configuration:
() = �� () � = � { ()}= � { ()}
and for the right configuration:
=1
=1
() = � { ()}
=1
=1
hence equivalent. Both systems use one summer and multipliers, but the left configuration is simpler since it only uses { } once.
6.
The systems below have input {(), []} and corresponding output {(), []}. For each system, determine whether it is (i) memoryless, (ii) stable, (iii) causal, (iv) linear, and (v) time invariant. (a) () = cos�()� (b) () = −∞ ∫ () (c) [] = cos(2[ + 1]) + [] (d) [] = 2[] ⁄ 2
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 9
ENSC3015
Signals and Systems
Answers: Memory- Stable less (a) YES YES (b) NO NO
Causal
Linear
YES YES
NO YES
Timeinvariant YES ?
(c) NO (d) YES
NO YES
NO YES
YES NO
YES NO
Comment
cos�()�non-linear function of () infinite integration across time () unstable [ + 1]: non-causal; cos( ): non-linear The gain factor 2 varies with time and is unbounded.
NOTE: For (d) we have that {[ − ]} = 2[ − ] but [ − ] = 2( − )[ − ] hence not time-invariant. THINK: For (b) the solutions say it is time-invariant but is that correct?
7.
Is it possible for a time-variant system to be linear? Demonstrate this for the RC circuit below by examining the relationship between input 1 () and output 2 () where the resistance is time-varying.
Answers: By KCL we get: If:
and:
then by superposition:
2 () 1 () = () + 2 () 1 () = �
2 () = �
1, ()
=1
2, () 1, () = () + 2, () 2, ()
=1
Because of () the system is time-varying and because of the above it is linear, so it is a linear, time-varying system.
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 10
ENSC3015
Signals and Systems
(a) Any real-valued continuous-time signal may be expressed as: () = () + (), where () is the even ∞ () is the∞odd component. Show that: ∞ component and � 2 () = � 2 ()+ � −∞2 () −∞ −∞ (b) Let be a linear, time-invariant system, that is () = {()}. If () is periodic with period , show that () is periodic with the same period. Answers: (a)
8.
∞
� −∞
Thus:
∞
−∞ ∞
∞ ∞ 2 () 2( ) + 2 � () () � + � −∞ −∞ −∞
(−) (−) = ()[− ()] = − () () 0
∞
+ � () () = � () () � () () −∞
Hence:
−∞ ∞
= � [2 () + 2 () + 2 () ()]
= We have that:
∞
= � [ () + ()]2
2 ()
−∞ ∞
0
∞
+ � () () = � (−) (−) 0
∞
0 ∞
∞
= − � () () + � () () =0 0
∞
0
∞
� 2 ()= � 2 ()+ � 2 () −∞
−∞
−∞
(b) If () is periodic with period then () = ( + ), what can we say about ( + )? ( + ) = {( + )} = {()} = () where {( + )} = {()} because is time-invariant. Hence () is also periodic with period .
23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA
Page 11
ENSC3015
Signals and Systems
Tutorial 2 (Solutions) Continuous-Time Systems (a) Find the zero-input response of the following LTI system with initial conditions 0 (0− ) = 2 and 0 (0− ) = −1: 2 () + () () () + 5 + 6() = 2 response (b) Find the zero-input of the following LTI system with initial conditions − − 0 (0 ) = 3 and 0 (0 ) = −4: 2 () + 4 () + 4() = () 2 Answers: 1.
(a) Using operational notation ≡ The characteristic equation:
, we rewrite: ( 2 + 5 + 6)() = ( + 1)() 2 + 5 + 6 = 0
( + 2)( + 3) = 0
is the factored :
to find the roots (or we can use the quadratic formula =
−5±�(5)2 −4(1)(6) 2(1)
1 = −2, 2 = −3 These are real, distinct roots so the solution is of the form: 0 () = 1 −2 + 2 −3 also: 0 () = −21 −2 − 32 −3 Using the initial conditions:
Hence:
0 (0− ) = 0 (0+ ) = 1 + 2 = 2 = 5 �→ 1 + − 0 (0 ) = 0 (0 ) = −21 − 32 = −1 2 = −3
(b) Using operational notation ≡
The characteristic equation: is the factored :
):
,
0 () = 5 −2 − 3 −3
>0
we rewrite: ( 2 + 4 + 4)() = () 2 + 4 + 4 = 0
( + 2)( + 2) = 0
1 = −2, 2 = −2 These are real, repeated roots so the solution is of the form: 0 () = 1 −2 + 2 −2 also (using product rule): 0 () = −21 −2 − 22 −2 + 2 −2 Using the initial conditions: 0 (0− ) = 0 (0+ ) = 1 = 3 = 3 �→ 1 + − 0 (0 ) = 0 (0 ) = −21 + 2 = −4 2 = 2 to find the roots:
Hence: 23/08/2019
Dr. Roberto Togneri, E&E Eng, UWA 0 () = 3 −2 + 2 −2 = (3 + 2) −2
>0
Page 12
ENSC3015 2.
Signals and Systems
Consider the continuous-time signals depicted below:
Evaluate the following convolution integrals: (a) () = () ∗ () *(b) () = () ∗ () *(c) () = () ∗ () Answers:
One solution method is the graphical explanation which can be depicted using a convolution animation tool. However, this is not reproducible with written tests/exams and in these written solutions. As such we need to find alternate methods, which are equally instructive in visualising and understanding signal operations and ∞ noting that convolution is commutative one integration. From 1 () ∗ 2 () =−∞∫ 1 ( )2 ( − and ) idea is to choose 2 () as the simpler waveform which is easier to reverse and shift and then to evaluate the integral in a piecewise fashion.
( − ) . Now: ) = ∫ ( ) (a) Let () = 2 () and () = 1 (); we have () = () ∗ ( −∞ 1 − 1 ≤ ≤ + 1 1 −1 ≤ − ≤=1� ( − ) = � 0 otherwise 0 otherwise so that: ∞
() = �
+1
()
−1
0 0 but () − ( − 1) is a finite duration signal for which we know the ROC is the entire s-plane. NOTE: When using ...
Similar Free PDFs
Tutorial Solutions
- 74 Pages
Module 1 - Tutorial Solutions
- 8 Pages
Tutorial 1 Solutions v2
- 5 Pages
Tutorial 2 Solutions
- 7 Pages
Statistics Tutorial 4 Solutions
- 7 Pages
Tutorial 9 Suggested Solutions
- 11 Pages
Tutorial 1 Solutions
- 4 Pages
Chapter 2 - Tutorial Solutions
- 6 Pages
Tutorial 3 Solutions
- 3 Pages
Tutorial 9 Solutions
- 3 Pages
Tutorial 7 Solutions
- 8 Pages
1 Tutorial Solutions
- 5 Pages
Tutorial K Solutions
- 2 Pages
Ch6 Tutorial Solutions Arens
- 7 Pages
Tutorial 3 solutions
- 5 Pages
tutorial 6 solutions
- 5 Pages
Popular Institutions
- Tinajero National High School - Annex
- Politeknik Caltex Riau
- Yokohama City University
- SGT University
- University of Al-Qadisiyah
- Divine Word College of Vigan
- Techniek College Rotterdam
- Universidade de Santiago
- Universiti Teknologi MARA Cawangan Johor Kampus Pasir Gudang
- Poltekkes Kemenkes Yogyakarta
- Baguio City National High School
- Colegio san marcos
- preparatoria uno
- Centro de Bachillerato Tecnológico Industrial y de Servicios No. 107
- Dalian Maritime University
- Quang Trung Secondary School
- Colegio Tecnológico en Informática
- Corporación Regional de Educación Superior
- Grupo CEDVA
- Dar Al Uloom University
- Centro de Estudios Preuniversitarios de la Universidad Nacional de Ingeniería
- 上智大学
- Aakash International School, Nuna Majara
- San Felipe Neri Catholic School
- Kang Chiao International School - New Taipei City
- Misamis Occidental National High School
- Institución Educativa Escuela Normal Juan Ladrilleros
- Kolehiyo ng Pantukan
- Batanes State College
- Instituto Continental
- Sekolah Menengah Kejuruan Kesehatan Kaltara (Tarakan)
- Colegio de La Inmaculada Concepcion - Cebu