Signals and Systems, Midterm 1 Solutions
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Transcript of Signals and Systems, Midterm 1 Solutions
ECE 203 Exam 1 Fall 2015 Discussion of Solutions
1) The complex number exp(-j 1.2 pi) has an angle between-pi and -(3/2) pi. So it lies in the 2nd quadrant, and theanswer is z2.
The complex number exp(j 1.8 pi) has an angle between1.5 pi and 2 pi. So it lies in the 4th quadrant, and theanswer is z4.
2) -6sin(4pi(t+1/7)) = 6cos(4pi t + (4/7)pi + pi/2) = 6cos(4pi t + (15/14)pi)
and so the phasor is 6exp(j(15/14)pi).An alternative correct answer can be obtained by usingthe principal angle, which can be computed by subtracting2 pi to get 6exp(-j(13/14)pi).
-7sin(3pi(t+1/5)) = 7cos(3pi t + (3/5)pi + pi/2) = 7cos(3pi t + (11/10)pi)
and so the phasor is 6exp(j(11/10)pi).An alternative correct answer can be obtained by usingthe principal angle, which can be computed by subtracting2 pi to get 6exp(-j(9/10)pi).
3) angle(3*exp(1j*(13/11)*pi))requires the principal angle. Since (11/13)pi > pi, wemust subtract 2pi to get -(9/11)pi or -2.5704.
angle(3*exp(-1j*(9/7)*pi))requires the principal angle. Since -(9/7)pi < -pi, wemust add 2pi to get (5/7)pi or 2.440.
4) In general, the statment
tt = t1:d:t2;
generates n numbers satisfying
tt = t1, t1+d, ..., t1+(n-1)d <= t2.
This implies (n-1) <= (t2-t2)/d, orn-1 = floor((t2-t2)/d), or n = 1+floor((t2-t1)/d).
With t1=-100, d=3/4, t2=88, we get1+floor(188/3*4) = 1+floor(250.67) = 1+250 = 251.
With t1=-100, d=3/4, t2=97, we get1+floor(197/3*4) = 1+floor(262.67) = 1+262 = 263.
5) If x(t) = sqrt(2) + sin(sqrt(12)t) - cos(sqrt(27)t),we see that sqrt(12) = sqrt(4*3) = 2 sqrt(3), andsqrt(27) = sqrt(9*3) = 3 sqrt(3).Hence, the radian frequencies are all multiples of sqrt(3).So the fundamental radian frequency is sqrt(3), and thefundamental period is 2 pi/sqrt(3) = 3.6276.
If x(t) = sqrt(7) + sin(sqrt(45)t) - cos(sqrt(80)t),we see that sqrt(45) = sqrt(9*5) = 3 sqrt(5), andsqrt(80) = sqrt(8*10) = sqrt(16*5) = 4 sqrt(5).Hence, the radian frequencies are all multiples of sqrt(5).So the fundamental radian frequency is sqrt(5), and thefundamental period is 2 pi/sqrt(5) = 2.8099.
6) Take the argument of the cosine, differentiate with respectto t, and divide by (2 pi). Answers aref_0 + mu 4 pi t + lambda exp(-2 pi lambda t)orf_0 + lambda 4 pi t + mu exp(-2 pi mu t).
7) f(A,B) = A/B. This problem was not graded; everyone gets 10 pts.You should read about varargin and nargin. They are very useful.
8) To be worked in class.
9) To be worked in class.