Fourier series for continuous and discrete time signals

Transcription

1 8-9 Signals and Systems Fall 5 Fourier series for continuous and discrete time signals The road to Fourier : Two weeks ago you saw that if we give a complex exponential as an input to a system, the output of the system is equal with the input multiplied with a value called the frequency response for the system. Mathematically, the previous sentence can expressed as follows:. in continuous time: We know h(t the impulse response of the system. We give the system the input signal x(t e jωt. We want to compute the output y(t. y(t h(t x(t, the convolution H(jω h(τx(t τdτ h(τe jω(t τ dτ h(τe jωt jωτ dτ h(τe jωt e jωτ dτ e jωt h(τe jωτ dτ e jωt H(jω x(th(jω h(τe jωτ dτ H(jω represents the frequency response of the system.. in discrete time: We know h[n] the impulse response of the system. We give the system the input signal x[n] e jωn. We want to compute the output y[n].

2 Fourier series for continuous and discrete time signals y[n] h[n] x[n], the convolution h[k]x[n k] k k k k h[k]e jω(n k h[k]e jωn jωk h[k]e jωn e jωk e jωn h[k]e jωk k e jωn H(e jω x[n]h(e jω H(e jω h[k]e jωk k H(e jω represents the frequency response of the system. Now lets add the fact that our system h(t is LTI. Let x(t a e jωt + a e jω3t + a 3 e jωt. The output will be the sum of the outputs for each of the complex exponentials. We can express the input signal x(t as a sum of 3 signals x (t, x (t and x 3 (t. x(t a e jωt + a e jω3t + a 3 e jωt x (t a e jωt x (t a e jω3t x 3 (t a 3 e jωt We can compute the output y(t taking into account that the system is LTI. y(t h(t x(t y(t h(t x (t + h(t x (t + h(t x 3 (t, h(t is an LTI system y(t y (t + y (t + y 3 (t

3 Fourier series for continuous and discrete time signals 3 The output signal y(t is going to be equal with the sum of y (t, y (t, y 3 (t. y (t h(t x (t, x (t is a complex exponential x (th(jω y (t h(t x (t, x (t is a complex exponential x (th(j3ω y 3 (t h(t x 3 (t, x 3 (t is a complex exponential x 3 (th(jω y(t y (t + y (t + y 3 (t x (th(jω + x (th(j3ω + x 3 (th(jω As you can see if we have complex exponentials as inputs the convolution can be simplified. You just have to multiply the input with the frequency response (which you still have to compute. Wait!!! Not all signals are complex exponentials, i.e. x[n] ( n u[n] is not a complex exponential. Oops. Fourier came up with a method of expressing signals as a linear combination of complex exponentials. So far in class we have tackled Fourier series for continuous time and discrete time, where the signal is periodic. A signal is periodic:. in continuous time: x(t is periodic if there exists a number T > such that x(t + T x(t. T is called the fundamental period and represents the smallest number greater than for which x(t + T x(t.. in discrete time: x[n] is periodic if there exists a integer number N > such that x[n + N] x[n]. N is called the fundamental period and represents the smallest number greater than for which x[n + N ] x[n]. In the next couple of pages we will see some examples for computing the Fourier series of continuous and discrete time signals.

4 4 Fourier series for continuous and discrete time signals Fourier Series for continuous time periodic signals The signal x(t is periodic with period T. The fundamental period is T and the fundamental frequency is ω o T. The signal x(t will be expressed as a linear combination of complex exponentials (harmonics, there are an infinite number of harmonics for the continuous case. This is the synthesis equation: x(t k X[k]e jkω t k jk t T X[k]e The coefficients X[k] are called the frequency domain representations of the signal x(t. This is the analysis equation: X[k] x(te jkωt dt jk T x(te t dt T T t<t > t<t >

5 Fourier series for continuous and discrete time signals 5. Problem You are given given the following signal. Compute the Fourier series for the signal. x(t cos(πt + 5 Solution: At the very beginning you have to determine the fundamental period of the signal T and the fundamental frequency ω. For x(t cos(πt + 5, T and w π. Method : We need to express the signal x(t as a linear combination of complex exponetials. In order to do that, we need to determine the coefficients X[k]. Therefore we will use the analysis equation. X[k] T t<t > x(te jkω t dt T t<t > jk T x(te t dt X[k] 4 x(te jkω t dt cos(πt + 5 e jkω t dt (ej(πt+ ( e j(πt+ 5 e jkωt dt + 4 ( e j ( e j 5 ( e j e j(πt+ 5 e jkω t dt e j(πt kω t dt + e j 5 e j(πt kπt dt + e j 5 e j(π kπt dt + e j 5 e j(πt+ 5 e jkωt dt e j(πt+kωt dt e j(πt+kπt dt e j(π+kπt dt

6 6 Fourier series for continuous and discrete time signals (a if k X[] 4 e jt dt (b if k 4 4 X[] 4 X[ ] 4 e jt dt ( e j 5 e jt dt + e j 5 ( e j 5 dt + e j 5 ( e j 5 + e j 5 e jt dt j cos(tdt j (e j 5 + e j 5 ej X[ ] 4 ( e j 5 ( e j 5 ( e j 5 + j cos(tdt + j e jt dt + e j 5 e jt dt + e j 5 (e j 5 + e j 5 e j 5 (c if k and k X[k] 4 4 e jt dt e jt dt sin(tdt, integration of sin and cos over periods e jt dt + e j 5 ( e j 5 ( e j 5 e jt dt dt sin(tdt, integration of sin and cos over periods e j(π kπt dt + e j 5 e jπ(k t dt + e j 5 e j(π+kπt dt e jπ(k+t dt

7 Fourier series for continuous and discrete time signals 7 f (t e jπ(k t f (t + e jπ(k (t+ e jπ(k t jπ(k e jπ(k t f (t e jπ(k+t f (t + e jπ(k+(t+ e jπ(k t dt e jπ(k+t jπ(k+ e jπ(k+t e jπ(k+t dt This implies that for k and k, X[k]. Method : The second method is by inspection. We want to compute the FS of the signal. We start from the signal and go backwards to the complex exponentials. x(t cos(πt + 5 (e j(πt+ 5 + e j(πt+ 5 ejπt e j 5 + e jπt e j 5 e j 5 e jπt + ej 5 e jπt Now we use the synthesis equation and pattern match. x(t k X[k]e jkπt x(t k X[k]e jkω t k jk t T X[k]e... + X[ ]e jπt + X[ ]e jπt + X[]e jπt + X[]e jπt + X[]e jt +... x(t e j 5 e jπt + ej 5 e jπt

8 8 Fourier series for continuous and discrete time signals From this we can see that X[ ] e j 5 and X[] ej 5. The rest are. Statement If the signal is a sin, cos or a linear combination of sin and/or cos then you can use the second method where you decompose the sin and the cos as complex exponentials. It may scrape off a lot off minutes at exams :.

9 Fourier series for continuous and discrete time signals 9. Problem Compute the Fourier series for the following signals. (a (b Solution: x(t cos(πt + sin(πt x(t cos(πt + sin(t (a At the very beginning you have to determine the fundamental period of the signal T and the fundamental frequency ω. For x(t cos(πt + sin(πt, T and w π. x(t cos(πt + sin(πt ( e jπt + e jπt + ( e jπt e jπt j ( e jπt + e jπt j ( e jπt e jπt ( jejπt + ( + je jπt ( + je jπt + ( jejπt Now we use the synthesis equation and pattern match. x(t k X[k]e jkω t k jk t T X[k]e x(t k X[k]e jkπt... + X[ ]e jπt + X[ ]e jπt + X[]e jπt + X[]e jπt + X[]e jt +... x(t ( + je jπt + ( jejπt From this we can see that X[ ] ( + j and X[] ( j. The rest are. This is a short solution. Think about the other method where you use integration.

10 Fourier series for continuous and discrete time signals (b At the very beginning you have to determine the fundamental period of the signal T and the fundamental frequency ω. For x(t cos(πt + sin(t, T and w π. x(t cos(πt + sin(t ( e jπt + e jπt + ( e jt e jt j ( e jπt + e jπt j ( e jt e jt j e jt + e jπt + ejπt j ejt Now we use the synthesis equation and pattern match. x(t k X[k]e jkω t k jk t T X[k]e x(t X[k]e jkπt k... + X[ ]e jπt + X[ ]e jπt + X[]e jπt + X[]e jπt + X[]e jt +... x(t j e jt + e jπt + ejπt j ejt From this we can see that X[ ] j, X[ ], X[] and X[] j. The rest are. This is a short solution. Think about the other method where you use integration.

11 Fourier series for continuous and discrete time signals 3. Problem 3 You are given the following continuous time signal. x(t δ(t 3k k x(t t Compute the FS for this signal. Solution: In this case we cannot apply the inspection method. The signal is not a sin, cos or linear combination of sin and/or cos. We have to apply the first method where we make use of the analysis equation to compute the coefficients. X[k] T t<t > x(te jkω t dt T t<t > jk T x(te t dt First we have to determine the fundamental period and the fundamental frequency. The fundamental period T 3 and the fundamental frequency is ω 3. X[k] e jk 3 δ(t e jk 3 t dt, use the sifting property of the δ function X[k] 3 X[k] k 3

12 Fourier series for continuous and discrete time signals 4. Problem 4 You are given the following continuous time signal. x(t t Compute the FS for this signal. Solution: In this case we cannot apply the inspection method. The signal is not a sin, cos or linear combination of sin and/or cos. We have to apply the first method where we make use of the analysis equation to compute the coefficients. X[k] T t<t > x(te jkω t dt T t<t > jk T x(te t dt First we have to determine the fundamental period and the fundamental frequency. The fundamental period T 4 and the fundamental frequency is ω π. 4 X[k] x(te j π kt dt 4 ( 4 x(te j π kt dt + x(te j π kt dt 4 ( 4 e j π kt dt e j π kt dt 4 e j π kt dt e j π kt j π k e j π k j π k

13 Fourier series for continuous and discrete time signals 3 4 e jπk j πk, e jπk cos(kπ j sin(kπ and k Z ( k j π k ( k j π k e j π kt dt e j π kt j π k 4 e j π k4 e j π k j π k e jk e jπk j π k e jπk j π k e jπk j πk, e jπk cos(kπ j sin(kπ and k Z ( k j πk X[k] ( ( k 4 j πk ( k j πk ( k ( k + 4j π k ( k jπk ( k jπk

14 4 Fourier series for continuous and discrete time signals 5. Problem 5 You are given the following continuous time signal. x(t t Compute the FS for this signal. Solution: In this case we cannot apply the inspection method. The signal is not a sin, cos or linear combination of sin and/or cos. We have to apply the first method where we make use of the analysis equation to compute the coefficients. X[k] T t<t > x(te jkω t dt T t<t > jk T x(te t dt First we have to determine the fundamental period and the fundamental frequency. The fundamental period T and the fundamental frequency is ω π. X[k] e jπkt dt e jπkt jπk e jπk jπk x(te jπkt dt ( te jπkt dt e jπkt dt te jπkt dt e jπk, e jπk cos(kπ j sin(kπ and k Z jπk ( k jπk

15 Fourier series for continuous and discrete time signals 5 Integration by parts. te jπkt dt integration by parts Let f(t and g(t to functions. We want to compute a b f(tg (tdt. We know that: This implies: (f(tg(t f (tg(t + f(tg (t f(tg (t (f(tg(t f (tg(t a b f(tg (tdt a b a b ((f(tg(t dt f (tg(t dt (f(tg(t dt f(tg(t a b a Coming back to our problem. We want to compute: b a b f (tg(tdt f (tg(tdt te jπkt dt f(t t f (t g (t e jπkt g(t e jπkt jπk te jπkt dt t e jπkt jπk e jπk jπk + jπk e jπkt jπk dt e jπkt dt e jπk jπk + e jπk, e jπk cos(kπ j sin(kπ and k Z jπk jπk ( k jπk + ( k jπk jπk ( k j πk ( k π k

16 6 Fourier series for continuous and discrete time signals ( ( k ( k j πk X[k] jπk ( j ( k j ( k j jπk πk ( j + ( k j ( k j jπk πk ( j πk + ( k j ( k j πk πk ( jπk π k + ( k π k jπk + ( k k + ( k π k + ( k π k + ( k π k + ( k π k

17 Fourier series for continuous and discrete time signals 7 Discrete Time Fourier Series for discrete time periodic signals The signal x[n] is periodic with period N. The fundamental period is N and the fundamental frequency is Ω o N. The signal x[n] will be expressed as a linear combination of complex exponentials (harmonics, in this case the number of harmonics is finite and equal with N. This is the synthesis equation: x[n] k<n > X[k]e jkω n k<n> jk n N X[k]e The coefficients X[k] are called the frequency domain representations of the signal x[n]. This is the analysis equation: X[k] N n<n > x[n]e jkω n N n<n > jk n N x[n]e We know that e jkω n is periodic N for the variable n. e jkω (n+n e jkω n e jkω N e jkωn jk N N e e jkωn jk N N e e jkω n e jk e jkω n e jkω n x[n] x[n + N ] k<n > k<n > k<n > x[n] X[k]e jkω n X[k]e jkω (n+n X[k]e jkω n This means that the signal x[n] is periodic with period N, which we already know.

18 8 Fourier series for continuous and discrete time signals We know that e jkω n is periodic N for the variable k. e j(k+n Ω n e jkω n e jn Ω n e jkω n e jn N n e jkω n e jn e jkω n e jkω n X[k] N X[k + N ] N n<n > n<n > N X[k] n<n > x[n]e jkω n x[n]e j(k+n Ω n x[n]e jkω n This means that the frequency representation X[k] is periodic with period N. x[n] is periodic with fundamental period N. Also X[k] is periodic with fundamental period N. That s why the number of harmonics is finite. Moreover, for e jkω n or e jkω n we can add or subtract to the phase n, without modifying the result. This will be useful in the inspection method for computing the DTFS. We will see it in action in a couple of minutes. e jkω n+n e jkω n e jn e jkω n e jkωn+n jk n+n N e e j( k N +n e j N k n N e jkωn e j N k n N e jkω n n e jkω n e jn e jkω n e jkωn n jk n n N e e j( k N +n e j k+n n N e jkωn e j k+n n N

19 Fourier series for continuous and discrete time signals 9 e jkω n+n e jkω n e jn e jkω n e jkωn+n jk n+n N e e j( k N +n e j k+n n N e jkωn e j k+n n N e jkω n n e jkω n e jn e jkω n e jkωn n jk n n N e e j( k N +n e j N k n N e jkωn e j N k n N

20 Fourier series for continuous and discrete time signals. Problem Compute the Discrete Time Fourier Series for the following signals. Solution: x[n] cos( π n + sin(π n At the very beginning you have to determine the fundamental period of the signal T and the fundamental frequency Ω. For x[n] cos( π n + sin( π n, N 4 and Ω π. x[n] cos( π n + sin(π n ( e j π n + e j π n + ( e j π n e j π n j ( e j π n + e j π n j ( e j π n e j π n ( π jej n + ( + π je j n ( + π je j n + ( π jej n Now we use the synthesis equation and pattern match. x[n] X[k]e jkωn X[k]e jk π n k<n > k<n> x[n] N k X[k]e jkω n 3 X[k]e jk π n k X[]e j π n + X[]e j π n + X[]e j π n + X[3]e j3 π n x[n] ( + je j π n + ( jej π n We do not have a match for e j π n. However, we can add jn at the phase and get something that we can match.

21 Fourier series for continuous and discrete time signals e j π n e j π n+n e j3 π n x[n] 3 X[k]e jk π n k X[]e j π n + X[]e j π n + X[]e j π n + X[3]e j3 π n x[n] ( + π jej3 n + ( π jej n From this we can see that X[] ( j and X[3] ( + j. The rest are. This is a short solution. Similar no the continuous case, one could have solved it using the analysis equation. Think how could one use that method no solve the problem. Would it be faster?

22 Fourier series for continuous and discrete time signals. Problem 3 You are given the following discrete time signal, where M < M and M is an even number. x[n] M M M M n Compute the DTFS for this signal. Solution: In this case we cannot apply the inspection method. The signal is not a sin, cos or linear combination of sin and/or cos. We have to apply the first method where we make use of the analysis equation to compute the coefficients. X[k] N n<n > x[n]e jkω n N n<n > jk n N x[n]e First we have to determine the fundamental period and the fundamental frequency. The fundamental period N M + and the fundamental frequency is ω. M+ X[k] M + M + M m M M M M + m M e jk n M M + n x[m]e jkω m M+ m, variable change n m + M m n M e jk M+ (n M e jk M+ n jk e M+ M

23 Fourier series for continuous and discrete time signals 3 M + ejk M + ejk M + ejk M + ejk M + M + M + M M+ M M+ M n e jk M+ n, the sum is a geometric progression e jk M+ (M + M+ M e jk M+ M+ (M M + jk e M+ M + M jk + M jk + (e M+ e M+ e jk π M+ (e jk + (e e jkπ M + M+ e jkπ M + M+ e jk π M+ e jk j sin(kπ M + j sin(k sin(kπ M + sin(k M+ π M+ M+ π M+ π M+ jk M+ (e jk π When k use l Hospital to get the value of X[]. π M+ e jk π M+ M + M jk + e M+ M+ e jk π M+ X[k] M + sin(kπ M + sin(k M+ π M+ Lets plot the signal x[n] and X[k] for M and M o {, 3, 6, 9}. (a M x[n] n

24 4 Fourier series for continuous and discrete time signals X[k] k (b M 3 x[n] n X[k] k (c M 6

25 Fourier series for continuous and discrete time signals 5 x[n] n X[k] k (d M 9 x[n] n

26 6 Fourier series for continuous and discrete time signals X[k] k