Signal Processing Assignment Help

Signal Processing Continuous and Discrete

For any help regarding Matlab Assignment Help visit : https://www.matlabassignmentexperts.com/, Email - info@matlabassignmentexperts.com or call us at - +1 678 648 4277 Matlab Assignment Experts

Problem

Problem 1:

A waveform f (t) with a real even spectrum F (jΩ) F (j

-5 0 0

0

-1 0

5 0 0

10

0



(ra d /s

is filtered by an ideal band-pass filter with a purely real frequency response H(jΩ) H ( j

-2 0 0

-4 0 0

2 0 0 4 0 0



(ra d /s

Determine the filter response y(t). Is this a causal filter?

Problem 2:

Find the impulse response of an ideal high-pass filter, with cut-off frequency Ωc: H ( j







c

c



( r ad / s

1 Matlab Assignment Experts

Hints: Feel free to do this any way you wish, but you may find it useful to consider H(jΩ) as a superposition of functions with known F T relationships. In particular, the solution of Prob. 5 in PS 1 might be useful and/or the following bits and pieces from the Frequency Domain class handout: the Fourier transform of a unit step function (p. 32), the duality property, the time shift property, time reversal property, linearity property, all might help you.

Problem 3: Use the Fourier transform of a sinusoid (p. 34 of the class handout) to express the Fourier series representation of a periodic waveform x(t) =

Σ∞

A sinn (nΩ t +0 φ )

n=0

n

as a Fourier transform.

Problem 4: Gibb’s phenomenon is associated with the synthesis of periodic waveforms with sharp (jump) discontinuities using a truncated Fourier series. Consider a periodic waveform x p (t) and its Fourier series Σ∞

x p (t) = X and let

Xn e

1 � T /2

=

n

j n Ω t0

n = −∞

x

T

(t) =

Þ p,N

x p(t)e

−jnΩ t

0

dt

−T /2

ΣN n = −N

X n e jnΩ0 t

be an approximation generated by truncating the series. It was observed in the late 1800’s tha t a finite series approximation created a ripple in the synthesized waveform in the region of a jump discontinuity in x p (t), and tha t although the width of the ripple decreased as more terms were included, the amplitude remained constant. The phenomenon was found to be Þ p,N

N = 15, 31, and 63. present for any finite N . The figure below shows the synthesis of a square wave x

(a ) N

= 1 5

(b ) N

= 3 1

(t) with

(c ) N

= 6

Gibb’s phenomenon was the subject of much discussion in the mathematical literature around 1900, and in fact was cited as a reason why Fourier analysis/synthesis did not work! The great mathematicians of the time did not understand what was going on. You, however, do understand it and your task is to explain the phenomenon.

2 Matlab Assignment Experts

(a) Explain the truncation of the series in the frequency domain as an ideal multiplicative filtering operation, and define the filter pass-band. (b) What is the equivalent time domain operation? OK - you win - it’s convolution, but the question is what are the two functions being convolved? Write the appropriate convolution integral. (c) Explain (i) why the height of the ripple on the square wave is constant for any finite series, and (ii) why the oscillatory frequency increases as N increases.

Problem 5: An “all-pass” filter has a frequency response magnitude |H(jΩ)| tha t is a constant (ie is independent of frequency Ω). (a) Show that the first-order system with the pole-zero plot below is an all-pass filter. j -

s - p Ia n

(b)

Show that any system with (i) the same number of poles and zeros, (ii) with all poles in the left-half s-plane, and (iii) the zeros mirror the poles about the imaginary axis, that is z i = −pi where x denotes the complex conjugate, is an allpass filter.

(c)

Use MATLAB to plot (i) the frequency response, (ii) the impulse response, and (iii) the step response of the filter with a pole and zero shown in (a) above. Choose an arbitrary gain constant.

(Some useful functions might be z p k ( ) , t f ( ) , f r e q s ( ) , b o d e ( ) , i m p u l s e ( ) , s t e p ( ) . )

(d) What useful purpose might an all-pass filter serve?

3 Matlab Assignment Experts

S o l u t i o n of P r o b l e m

P r o b l e m 1: Y (jΩ) = H(jΩ)F (jΩ) and it is shown in the below figure Y ( jw) 10

y(t) =

1 �∞ Y (j Ω)ej Ωt dt 2π −∞ −�200 10 �400 jΩt 2π = ( 200e dt + ejΩt dt) 10 = 2πj t −400 400j t 200j t (e −−e sin(200t)) − e−400j t + e−200j t ) = 10(sin(400t) πt

It is not a casual filter (since y(t) = h(t)/5 and h(t) =/ 0 for some t < 0).

P r o b l e m 2: We define H∗(jΩ) = 1 − H(jΩ). Then H∗(jΩ) is a low pas filter, matching Prob. 5 in PS 1, which we have already found it’s impulse response: H (j Ω) = 1 − H ∗(j Ω) F −1 (H (j Ω)) = F −1 (1) sin(Ω − Fc t)−1 (H ∗(j Ω)) h(t) = δ(t) − πt h(t) = δ(t) − h ∗(t) Matlab Assignment Experts

P r o b l e m 3: Σ∞

Σ∞

x(t) = Σ∞

X (j Ω) =

A n sin(nΩ0 t + φn ) =

A n (sin(nΩ0 t) cos φn + cos(nΩ0 t) sin φn )

n=0 n= 0

n=0

A∞n ( F (sin(nΩ0 t)) cos φn + F (cos(nΩ0 t)) sin φn )

Σ

n= 0

∞ − φnΩ + 0nΩ ))) +cossinφnφ+n (π (δ(Ω nΩ0+ ) +nΩ δ(Ω X(jΩ) = −jπ A n (cos A nΣφ ((cos φn (δ(Ω + j sin − nΩ ) + 0(− j sin φ n )−(δ(Ω n (−jπ 0) − δ(Ω n ) (δ(Ω 0)) + nΩ0 )))) n= 0 � � X (j Ω) = −j π A nΣ∞ej φn δ(Ω − nΩ0 ) − e−j φn δ(Ω + nΩ0 ) n=0

P r o b l e m 4: ( a ) TheΩideal multiplicative filtering operation is a low c = N Ω0 : � pass filtering with the pass-band 1 |n| ≤ N , |Ω| ≤ Ωc H n= 0 |n| > N , |Ω| > Ωc X�n = X n H n � T2 1 x�(t) = x (t) ⊗ h(t) = x(τ )h(t − τ )dτ T — T2 We can prove that why convolution is in this specific integral form for our Periodic Exponential Fourier Transform:

( b ) It ’s a convolut ion in t his specific form:

x�(t) =

� Σ∞ X n H ne

n=−∞

=jnΩ

0 Σ∞ � t

� n=−∞

2 � � x�(t) = 1 � T h(τ ) Σ∞ X n e −jnΩ n=−∞ T −T 2 � T2 x�(t) = 1 h(τ )x (t − τ )dτ T — T2 We can also find the specific form of our filter:

h(t) =

Σ∞ n=−∞

H n e jnΩ0 t =

�

� � � T2 1 −jnΩ τ 0 0 h(τ )e dτ e jnΩ t Xn T �— T2 � 1 � T2 � h(τ ) Σ∞ � dτ = Xn e τ 0 jnΩ t 0 e n=−∞ T −T 2

ΣN

ΣN n=−N

e

j nΩ0 t

= 1+ 2 n=1

cos (nΩ0 t) = −1 + 2

ΣN

cos (nΩ0 t) n=0

Matlab Assignment Experts

� 0

j nΩ (t−τ

� )

dτ

Here we use below trigonometric relation to simplify our summation, where ϕ = 0 and α = Ω0t: (n+ 1)α 2 sin � · cos � (ϕ + cos ϕ + cos (ϕ + α ) + cos (ϕ + 2α) + · · · + cos (ϕ + nα) = α sin 2 � (N + 1)Ω t � 0 sin · cos ( N Ω02t ) 2 ΣN sin Ω0 t2 h(t) = −1 + 2 cos (nΩ0 t) = −1 + 2 Then we use below trigonometric n = 0 relation to simplify our impulse response: h(t) = −1 + 2

2 sin θ cos − ϕ) � ϕ = sin(θ � + ϕ ) + sin(θ N Ω 0t sin · cos ( ) (N + 1)Ω0 t 2 = −1 + 2 sin Ω0t 2 � sin (N + 1 )Ω0 t 2 Ω0t h(t) = sin 2

� sin (N + �

1

� )Ω2 t +0 sin ( Ω0 t ) sin Ω0 t2

nα

)2

2

Here we further analyze this h(t) function, so we can use its properties for the next part of the problem: Note that for small t values, h(t) can be approximated to: � � sin (N + 1 )Ω0 t 2 Ω0t t → 0 ⇒ h(t) → 2

The value of h(t) |t→0 , for large N values, is very close to h∗ = s i n ( N Ω t ) which can be πt 0 obtained for a Continuous Fourier Transform of a Low Pass Filter with Ωc = NΩ0. Hence, in the vicinity of t = 0, h(t) acts like a sinc function (with the maximum value of 2N + 1), but as soon as it gets close to its boundaries (|t| ≤ T ⇒ |Ω0t| ≤ π), it 2 oscillates quickly with Ω = (N + 0.5)Ω0 around −1 and +1.

Note that forbetween the Continuous we expect a h∗ (t); it is 1 be evaluated −∞ to +Fourier ∞ and Transform, contained with an envelope in where the form of to .

On the other hand, for our case of Periodic Fourier Transform, h(t) it is to be evaluated between − periodic.

2

T

t

to + T and is also

2

Note that for any N value, if x(t) ≡ 1 then ≡ 1 which that following relation holds for any N: � x�(t) T2 � means T 2 1 2 1= h(τ )dτ = h(τ )dτ 0 T — T2 T Furthermore, for very large N values, the h(t) function become very narrow, and hence above integral can be approximated to the integration around any finite, non-zero interval around t = 0: (Eq. 1)

N → ∞ ⇒ ∀t s.t . 0 ∗< t ≤ :

∗

T 2

1 T

� 0

t∗

h(τ )dτ → � 1T T

2

0

h(τ )dτ

Matlab Assignment Experts

This properties can be used to prove that x�(t) values converge to x(t) values at any continuous point and to the mean of right and left limits at any stepwise discontinuous point of x(t). Furthermore, the structure of h(t) shows tha t the ripple will vanish at any point in which x(t) is continuous. Besides, as we will see in the next part, only a finite and determined value of ripple will be allowed to remain and it will be pushed to the edge of discontinuities. ( c ) The output signal is a convolution of the original signal with the h(t). Hence, it is very clear that ripples are due to deviation of h(t) from ideal case of δ(t). Consequently, as soon as N goes up, h(t) becomes more like a δ(t) function, acts more locally, and the output signal ripples with an increases frequency.

For very large N values, the ripple percentage is a fixed amount, although its location is dependent on N value. The amount of ripple is fixed, because although different h(t) functions have different curves, but always the maximum ripple corresponds to the point where the edge of the central lobe matches with the discontinuity. We will prove this rigorously, but in fact since the ratio of the area of the central lobe, to the rest of the lobes remains constant, the ripple percentage remains constant as well.

Note that due to Eq. 1, only the behavior of function around discontinuity mat ters (as long as discontinuities have a finite non-zero distance from each other). Hence, without the loss of generality, we extend discontinuity limits to the the full extent of the function and consider our function in the below form where A value corresponds to the discontinuity � jump: 2 x(t) = T A 02 ≤ t < T Below figure shows this conditions: 0 − ≤t<0 A

x(t), to the scale

t_max

h(t) for N=10, not to the scale

0

-T/2









 0



 T/2

By moving h(t) on the x(t), we can realize that, at each point, the x�(t) is an average of neighborhood points.

Particularly, the sign of lobes determine that whether those Matlab Assignment Experts

point have a positive or negative contribution. Since, major contribution comes from the central lobes, we can guess that the maximum ripple occurs when the edge of cen tral lobe match with t = 0. In that case, the next negative lobe cannot decrease the x�(t) value and we have a maximum ripple. The next maxima and minima of ripples also correspond to other lobe edges touching t = 0.

Now we prove this rigorously. For our specific case of x(t), and our even h(t) function, we can simplify the convolution integral: � T2 � T A � T −t 2 A 2 x�(t) = 1 h(t − τ )x(τ )dτ = h(t − τ )dτ = h(τ )dτ 0 −t T T T −T 2 dx�(t) = −h( T dt 2 − t) + h(t) dxb(t) We are interested in 0 < t << T , and hence 2 ≈ h(t). Consequently, maxi dt 1 mum/minimum values correspond to (N + )Ω0tmax/min = kπ2 such that 0 < k << N .

1 π + 2 )Ω 2(N + 21) The highest maximum corresponds to the t = = max T . (N Now we0 compute the x�(tmax) value by breaking the original integral to two parts: � T −t m ax 0 A � 2T −t m ax h(τ )dτ = A � 1 1+ 2 ( h(τ )dτ h(τ )dτ ) x�(t max ) = −t −t 0 max max T T T Now note that for large N values from Eq. 1 we can conclude that: � T −t m ax � T 2 1 1 1 2 h(τ )dτ ) = h(τ )dτ ) = 0 0 T T 2 Also for the other part of integral, τ is very close to zero and we can simplify h(t) with a sinc form and also change integration variable by θ = (N + 1 )Ω0t: 2 � T

1

−t 0

m ax

h(τ )dτ ) =

� T

1

−t 0

sin 2((N0 + m ax

1 )Ω τ ) dτ Ω0 τ =

2

� T1

sin θ 2 −π θ Ω0 0

dθ =

� π1

π

0

sin θθ

dθ

A short survey of literature or a Taylor expansion of s i n θ ends to: θ � 1π 0π sin θ θ dθ1= + 2 0.089490...

Hence the maximum ripple corresponds to: 1 1 x�(tmax) = A( + + 0.089490...) = A(1 + 0.089490...)

2

2

This corresponds to about 9% overshoot and this overshoot is only dependent on the local discontinuity jump and independent of N or specific form of x(t). Other proofs, for less general cases, could also be found in the Wikis.

Matlab Assignment Experts

P r o b l e m 5: (a)

s−3 � s + 3� � j Ω − 3 � |H (j Ω)| = |K | � � j Ω + 3 � = |K | H(s) = K

( b ) Consider this pole-zero plot:

�

j

r1 r r

s -p I a n q q

2

q 3

Let the vectors from the poles and zeros to an arbitrary test frequency Ω be r i and qi: |q | |q | |q |

2 3 |H (j Ω)| = |K | 1 = |K | |r 1 | |r 1 | |r 1 |

( c)

Msince AT LA m aall n di − lin e : of the system order. Therefore, this given pole-zero configuration, as well as all |qi|B= C|rom regardless i| for who satisfy problem conditions, are all-pass filters. >> H_s =tf([1 - 3 ] , [ 1 3 ] ) Tr a n s f e r f u n c t i o n :

s - 3

s + 3 >> bode(H_s); >> s u b p l o t ( 2 , 1 , 1 ) Matlab Assignment Experts

>> s t e p ( H _ s ) ; >> s u b p l o t ( 2 , 1 , 2 ) >> impuls e(H_s); Bode Diagram

Magnitude (abs)

1 0.8 0.6 0.4

Phase (deg)

0.2 0 180 135 0 10

−1

10 90

0

10

Frequency (rad/sec)

1

10

2

45 Step Response

Amplit ude

1 0.5 0 −0.50

0.2

0.4

0.6

−1

1 Time (sec)

1.2

1.4

1.6

1.8

2

1.4

1.6

1.8

2

Impulse Response

0 Amplit ude

0.8

−2 −4 −6 −80

0.2

0.4

0.6

0.8

1 Time (sec)

1.2

−10

( d ) These filters are useful for manipulating the phase of the spectral components in a signal, without altering its magnitude.

Matlab Assignment Experts