Assignment Solutions Networks and systems August 8, 7. Consider an LTI system with transfer function H(jw) = input is sin(t + π 4 ), what is the output? +jw. If the Solution : C For an LTI system with frequency response H(jw) and input A sin(ω t+θ) the output is given by A H(jω ) sin(ω t + θ + H(jω )). H(jω ) = + = H(jω )) = π 4 Output = sin(t + π 4 π 4 ) = sin(t). In the circuit shown below, v s (t) is the input and v c (t) the output. All initial conditions are zero. If v s (t) = cos(πt+ π 4 ) sin( πt π 6 ), find the steady-state value of v c (t) Solution : B
For an LTI system with frequency response H(jw) and input A sin(ω t+θ) the output is given by A H(jω ) sin(ω t + θ + H(jω )). Similarly input A cos(ω t + θ) gives A H(jω ) cos(ω t + θ + H(jω )). H(jω) = jωc R + jωc = + jωcr (a) When input = cos(πt + π 4 ) ω = π H(jω ) = + j π. = + j = π H(jω ) = π 4 Output = cos(πt + π 4 π 4 ) (b) When Input = sin( πt π 6 ) ω = π H(jω ) = + j π. = + j = π H(jω ) = π Output = sin( πt π 6 π ) Total output : = cos(πt + π 4 π 4 ) sin( πt π 6 π ) = 5 cos(πt) + cos( πt). If the signal x(t) = +4 cos(ω t π 6 )+ cos(ω t+ π ) is written as Fourier series in the form a + n= [a n cos(nω t) + b n sin(nω t)], find the value of a + a a + b + b + b. Solution : A cos(ω t π 6 ) = cos(ω t) + sin(ω t)
cos(ω t + π ) = cos(ω t) sin(ω t) x(t) = + 4[cos(ω t) + sin(ω t) ] + [cos(ω t) sin(ω t) x(t) = + cos(ω t) + cos(ω t) sin(ω t) + sin(ω t) Hence we have a = ; a = ; a = ; b = ; b = ; b = a + a a + b + b + b = ] 4. Evaluate the integral sin(mπt) sin(nπt)dt. Solution : A,D I = sin(mπt) sin(nπt)dt = cos((m n)πt) cos((m + n)πt) = [ (sin((m n)π) (m n)π ) ( sin((m + n)π) ) ] (m + n)π m = and n= = I = = [ sin(π) π sin(π) ] π m = and n= = I = [ π ] π = sin(ax) (Since lim = a) x x 5. Evaluate the integral sin(mπt) cos(nπt)dt Solution : A,C I = = = sin(mπt) cos(nπt)dt [ sin((m + n)πt) + sin((m n)πt) ( cos((m + n)π) (m + n)π ) ( cos((m n)π) ) ] + (m + n)π (m n)π (m n)π
m = and n= = m = and n= = I = [ ] = I = = [ cos(6π) π + cos(π) ] π cos(ax) sin (ax) (Since lim = lim = ) x x x x 6. Find the DC term a in the Fourier series expansion of the following signal. x(t) 5 4 4 5 t Solution : D Time Period = 4 DC term = a = T T x(t)dt T x(t)dt = Area of triangle = = Hence a = 4 7. For the following signal which of the following is/are true? x(t) 4 4 t Solution : A,B 4
(a) DC term is zero since signed area between - to is zero. (b) Since x(t) has half wave anti symmetry i.e. x(t + T ) = x(t) the even harmonics are zero. (c) Odd harmonics are clearly not zero. (d) Cosine terms are not zero because x(t) is NOT odd symmetric. 8. Identify the even-symmetric functions in the following: Solution : C Only option c is even symmetric 9. Identify the Odd-symmetric functions in the following: Solution : A,D Options a and d are odd symmetric. Let f(t) = f e (t) + f o (t) = e j(t+ π ) where fe (t) = f e ( t) (even part) and f o (t) = f o ( t) (odd part) Solution : B,D f(t) = e j(t+ π ) f(t) = cos(t + π ) + j sin(t + π ) f(t) = sin(t) + j cos(t) f( t) = sin(t) + j cos(t) f e (t) = f(t) + f( t) f e (t) = cos(t + π ) + j sin(t + π ) f(t) f( t) f o (t) = f o (t) = sin(t) 5
. Which of the following signals has only sine terms in the Fourier Series expansion? Solution : A,C (a) sin(t) cos(t) =.5 sin(t) and hence contains only sine terms.(same result also be concluded from observing that it has odd symmetry) (b) sin (t) = cos(t) and thus contains a cos term. (c) Possesses odd symmetry and hence contains only sine terms. (d) Does not possess odd symmetry and hence does NOT contain only sine terms.. Let x(t) = x(t + T ).If its FS expansion has only even-order harmonics, the fundamental period of the signal must be T / Solution : False Suppose If the fundamental period is T even then there are only even 4 order harmonics present and hence we cannot conclude that fundamental period is T Let us consider an illustrative example x(t) = sin(4t) + sin(8t) Consider T = π Clearly the condition x(t) = x(t + T ) is satisfied. With respect to T signal x(t) has only even order harmonics (in particular 4th and 8th). However the fundamental period of x(t) = T 4. Let x(t) = x(t + T ). One period of x(t) is shown below:select all the applicable properties in the Fourier Series expansion 6
A T 4 T T t A Solution : B,C The given signal has half wave anti symmetry i.e. x(t + T ) = x(t) and hence there are only odd harmonics in the signal Only cosine terms are present due to even symmetry of x(t). 4. Which of the following signals has only odd harmonics in the Fourier Series expansion? Recall that by harmonics we mean those frequencies that are integer multiples of the fundamental frequency. Solution : A,B,D (a) Clearly has only odd harmonics. (b) Even though 4 might look tempting to say even harmonic it is NOT because sin(4t) has only one component i.e. fundamental frequency and hence has only odd harmonics.(can also be concluded from the half wave anti symmetry of sin(4t) ). (c) f(t+t/) -f(t) i.e. does not have half wave anti symmetry and hence does NOT contain only odd harmonics. (d) f(t+t/) -f(t) i.e. does not have half wave anti symmetry and hence does NOT contain only odd harmonics. (e) f(t+t/) = -f(t) i.e. has have half wave anti symmetry and hence contains only odd harmonics. 5. For the waveform shown below, the amplitude and frequency are A = 7
5V, f = 5Hz.If the signal is represented as f(t) = a π + b π sin(πt) + c cos(πt) +..., what is the value of a + b + c π A T t Solution : - : (a) DC component a = T x(t)dt T = T ( π ) A sin t dt T T = A [ T T π cos = A π Hence a = A = ( π T T ) cos ( π )] T (b) Coefficient of sin(πt) Since the waveform is even symmetric the sine terms are zero. Hence b= (c) Coefficient of cos(πt) a = T x(t) cos(πt)dt T = T T ( π ) A sin t cos(πt)dt T = T A sin(5πt) cos(πt)dt T 8
= A T [sin(5πt) sin(5πt)]dt T [ = A (cos(5πt ) T 5π = A [ T 5π ] 5π = A T = π 5π ) ( cos(5π T ) ) ] 5π Hence c=- Therefore a+b+c= - 9