3. Frequency-Domain Analysis of Continuous- Time Signals and Systems

Transcription

1 3. Frequency-Domain Analysis of Continuous- ime Signals and Systems 3.. Definition of Continuous-ime Fourier Series ( ) 3.2. Properties of Continuous-ime Fourier Series (3.5) 3.3. Definition of Continuous-ime Fourier ransform ( ) 3.4. Properties of Continuous-ime Fourier ransform ( ) 3.5. Frequency Response (3.2, 3.8, 4.4) 3.6. Linear Constant-Coefficient Differential Equations (4.7)

2 3.. Definition of Continuous-ime Fourier Series A continuous-time signal x(t) with period can be represented by a continuous-time Fourier series, i.e., x(t) k X(k) exp j 2 kt, (3.) where X(k) is given by X(k) x(t) exp X(k) is called the spectrum of x(t). 2 j kt dt. (3.2) (3.) and (3.2) show that a continuous-time periodic signal can be decomposed into a set of continuous-time elementary signals. Any continuous-time elementary signal X(k)exp(j2kt/) is periodic, and has the frequency 2k/ and the coefficient X(k).

3 3... Derivation of Continuous-ime Fourier Series Assume that x(t) can be represented by (3.). We show that X(k) is given by (3.2). Substituting k for k in (3.), we obtain x(t) k X(k)exp 2 j kt. (3.3) Next, (3.3) is multiplied by exp(j2kt/), integrated over one period, and divided by. hat is, k x(t) exp X(k)exp 2 j kt dt 2 j kt exp 2 j kt dt. (3.4) Changing the order of the integration and the summation on the right side of (3.4), we obtain

4 Since k X(k) exp x(t) exp 2 j exp 2 j 2 j (k k)t dt kt dt (k k)tdt., 0, k k, k k the right side of (3.5) equals X(k), and (3.2) is derived Convergence of Continuous-ime Fourier Series (3.5) (3.6) he integral in (3.2) converges when the following conditions are satisfied. () In any period, x(t) is absolutely integrable. hat is, there exists a finite constant B such that

5 x(t) dt B. (3.7) (2) In any period, x(t) has a finite number of maxima and minima. (3) In any period, x(t) has a finite number of discontinuities, and has both the left-sided limit and the right-sided limit at each of these discontinuities. he above conditions are called the Dirichlet conditions. It should be noted that they are sufficient for the convergence of the integral in (3.2) but unnecessary. Assume that the Dirichlet conditions are satisfied, and the integral in (3.2) converges. hen, the series in (3.) converges but may not converge to x(t) everywhere. At a continuity, it converges to x(t). At a discontinuity, it converges to the average of the left-sided limit and the right-sided limit of x(t). Example. Determine the Fourier series coefficients for each of the

6 following signals: () x(t)=sin( 0 t). (2) x(t)=+sin( 0 t)+2cos( 0 t)+cos(2 0 t+/4). (3) Over period [/2, /2), x(t) is defined as (4) x(t) n x(t), 0, t t 0 t t 3.2. Properties of Continuous-ime Fourier Series Linearity (t n). Suppose that x (t) and x 2 (t) have the same period, and a and a 2 are two arbitrary constants. If x (t)x (k) and x 2 (t)x 2 (k), then 0.

7 Differentiation If x(t)x(k), then Shifting If x(t)x(k), then a x (t)+a 2 x 2 (t)a X (k)+a 2 X 2 (k). (3.8) dx(t) dt 2 j kx(k). (3.9) 2 x(t t 0) X(k) exp j kt 0, where t 0 is an arbitrary real number. If x(t)x(k), then 2 x(t) exp j k 0t X(k k 0), (3.0) (3.)

8 where k 0 is an arbitrary integer Scaling If x(t)x(k), then where a is a nonzero real number. When a>0, (3.2) becomes x(at)x[sgn(a)k], (3.2) x(at)x(k). (3.3) hus, only X(k) cannot specify x(t) uniquely. is also required. Letting a= in (3.2), we obtain x(t)x(k), (3.4) i.e., the reversal property of the continuous-time Fourier series. From (3.4), the following conclusions can be drawn:

9 () x(t) even X(k) even. (2) x(t) odd X(k) odd Conjugation If x(t)x(k), then x * (t)x * (k). (3.5) From (3.5), the following conclusions can be drawn: () Im[x(t)]=0 X(k)=X * (k). (2) Re[x(t)]=0 X(k)=X * (k). (3) Im[X(k)]=0 x(t)=x * (t). (4) Re[X(k)]=0 x(t)=x * (t) Convolution Assume that x (t) and x 2 (t) have the same period. If x (t)x (k)

10 and x 2 (t)x 2 (k), then x( )x 2(t )d X(k)X2 (k), x (t)x (3.6) where the integral is called the periodic convolution integral of x (t) and x 2 (t). Assume that x (t) and x 2 (t) have the same period. If x (t)x (k) and x 2 (t)x 2 (k), then (t) X (m)x (k m), (3.7) 2 m where the sum is called the convolution sum of X (k) and X 2 (k) Parseval s Equation 2 If x(t)x(k), then x(t) 2 dt X(k). (3.8) k 2

11 3.3. Definition of Continuous-ime Fourier ransform A continuous-time signal x(t) can be represented by a continuoustime Fourier integral, i.e., x(t) where X() is given by 2 X( ) X( )exp x(t) exp jt d, jt dt. (3.9) (3.20) (3.20) is called the continuous-time Fourier transform, and (3.9) is called the inverse continuous-time Fourier transform. X() is called the spectrum of x(t). From (3.9) and (3.20), we see that a continuous-time signal can be decomposed into a set of continuous-time elementary signals. Any continuous-time elementary signal X()exp(jt)d/(2) is periodic, and has the frequency and the coefficient X()d/(2).

12 3.3.. Derivation of Continuous-ime Fourier ransform Assume that x(t) is x(t) extended with period. hen, x(t) k x ( )exp j 2 kd exp j 2 kt. (3.2) Letting =2/, we obtain x (t) 2 2 k x ( )exp Letting 0, we obtain x(t) 2 x( )exp (3.23) shows that x(t) can be expressed as x(t) 2 jkd exp jkt. jd exp jt d. X( )exp jt d, (3.22) (3.23) (3.24)

13 where X( ) x(t) exp jt dt Convergence of Continuous-ime Fourier ransform (3.25) When the following conditions are satisfied, the integral in (3.20) converges. () Over (, ), x(t) is absolutely integrable. hat is, there exists a finite constant B such that x(t) dt B. (3.26) (2) Over (, ), x(t) has a finite number of maxima and minima. (3) Over (, ), x(t) has a finite number of discontinuities, and has both the left-sided limit and the right-sided limit at each of these discontinuities.

14 he above conditions are called the Dirichlet conditions. It should be noted that they are sufficient for the convergence of the integral in (3.20) but unnecessary. Assume that the Dirichlet conditions are satisfied, and the integral in (3.20) converges. hen, the integral in (3.9) converges but may not converge to x(t) everywhere. At the continuities, it converges to x(t), but at the discontinuities, it converges to the average of the leftsided limit and the right-sided limit of x(t) Examples of Continuous-ime Fourier ransform he continuous-time Fourier transform can be used to represent both aperiodic continuous-time signals and periodic continuous-time signals. First we consider aperiodic continuous-time signals. Example. Find the Fourier transforms of the following signals: () x(t)=(t).

15 (2) x(t), 0, t t 0 t t (3) x(t)=e at u(t), Re(a)<0. (4) x(t)=e at u(t), Re(a)>0. (5) x(t)=e a t, Re(a)<0. 0. Next we consider periodic continuous-time signals. Example. Prove exp(j 0 t)2( 0 ). (3.27) Example. Find the Fourier transform of x(t)=sin( 0 t). Example. Assume that x(t) is a continuous-time signal with period, and X(k) is the Fourier series coefficient of x(t). Determine X(), the Fourier transform of x(t).

16 3.4. Properties of Continuous-ime Fourier ransform Linearity If x (t)x () and x 2 (t)x 2 (), then a x (t)+a 2 x 2 (t)a X ()+a 2 X 2 (), (3.28) where a and a 2 are two arbitrary constants Differentiation If x(t)x(), then If x(t)x(), then dx(t) dt jx( ). (3.29) dx( ) jtx(t). (3.30) d

17 Example. Evaluate the derivative of x(t) at t=0. Assume that the Fourier transform of x(t) is given as follows: / 2, 0.5 () X( ), , j, (2) X( ) j, 0, 0 0 otherwise. Example. Let x(t)x(). Prove y(t) t x( )d Y( ) X( ), j X(0) (0), 0. 0 (3.3)

18 Example. Let x(t)x(). Prove y(t) Shifting If x(t)x(), then x(t), t 0 jt x(0) (0), t 0 where t 0 is an arbitrary real number. If x(t)x(), then where 0 is an arbitrary real number. Y( ) X( )d. (3.32) x(tt 0 )X()exp(jt 0 ), (3.33) x(t)exp(j 0 t)x( 0 ), (3.34) Example. A modulator is a basic unit in a communication system.

19 It converts a signal into another one which can be transmitted more effectively and more efficiently. A sinusoidal amplitude modulator is described as r(t)=s(t)p(t). (3.35) s(t) is called the modulating signal and bears the wanted information. Since it is a low-frequency signal, s(t) cannot be transmitted effectively and efficiently. p(t) is called the carrier signal. It is a highfrequency sinusoidal signal. Here, we assume p(t)=cos( 0 t). (3.36) r(t) is called the modulated signal. Since it is a high-frequency signal, r(t) can be transmitted effectively and efficiently. Let the spectrum of s(t) be S(). Find R(), the spectrum of r(t). Example. A demodulator recovers the modulating signal from the modulated signal. In the last example, s(t) can be recovered from r(t)

20 by a sinusoidal amplitude demodulator, which is characterized by g(t)=r(t)p(t) (3.37) followed by a low-pass filter. Find G(), the spectrum of g(t), and its expression after the low-pass filtering Scaling If x(t)x(), then x(at) where a is a nonzero real number. Letting a= in (3.38), we obtain X, (3.38) a a x(t)x(), (3.39) the reversal property of the continuous-time Fourier transform.

21 From (3.39), the following conclusions can be drawn: () x(t) even X() even. (2) x(t) odd X() odd Conjugation If x(t)x(), then x * (t)x * (). (3.40) From (3.40), the following conclusions can be drawn: () Im[x(t)]=0 X()=X * (). (2) Re[x(t)]=0 X()=X * (). (3) Im[X()]=0 x(t)=x * (t). (4) Re[X()]=0 x(t)=x * (t).

22 Symmetry If x(t)x(), then Proof. Substituting for t in X(t)2x(). (3.4) X( ) x(t) exp jt dt, (3.42) we obtain X( ) x( )exp j d. (3.43) Substituting t for in (3.43), we obtain X(t) jt d. x( )exp (3.44) Substituting for in (3.44), we obtain (3.4).

23 Example. Prove sin( t 0 t), 0, 0 0. (3.45) Example. Find the Fourier ransform of x(t)=2/(+t 2 ) Convolution If x (t)x () and x 2 (t)x 2 (), then x (t)x 2 (t)x ()X 2 (). (3.46) his property is proved as follows: x x ( ) ( )x x 2 2 (t )d exp (t )exp jt jt dt dtd

24 X 2 ( ) X ( )X 2 x ( ). ( )exp Example. Find x (t)x 2 (t), where j () x (t)=e at u(t), Re(a)<0, x 2 (t)=e bt u(t), Re(b)<0, and ab. sin( t) sin( 2t) ( 2) x(t) and x 2(t). t t If x (t)x () and x 2 (t)x 2 (), then Example. Let d (t)x 2(t) [X( ) X ( )]. 2 x 2 sin( t) sin( t / 2) x(t). 2 t (3.47) (3.48) (3.49)

25 Find the Fourier transform of x(t) Parseval s Equation If x(t)x(), then x(t) his property is proved as follows: 2 2 dt X( ) d. 2 (3.50) x(t) 2 dt x(t)x * (t)dt 2 2 x(t) X X 2 * * ( ) X( )exp ( )X( )d x(t) exp jt 2 d jt * dt dtd X( ) 2 d. (3.5)

26 Example. Evaluate the integral of x(t) 2 over (,). Assume that the Fourier transform of x(t) is given as follows: () X( ) (2) X( ) / 2, 0.5, , j, j, 0, 3.5. Frequency Response 0 0. otherwise A linear time-invariant continuous-time system can be described by the frequency response, which is defined as the Fourier transform of the impulse response.

27 3.5.. Response to exp(j 0 t) Let the input of a linear time-invariant continuous-time system be x(t)=exp(j 0 t). hen, the output of the system will be y(t)=exp(j 0 t)h( 0 ), (3.52) where H() is the frequency response of the system. Proof. Let h(t) be the impulse response of the system. hen, y(t) h( )exp exp( j exp( j 0 0 t) j t)h( ). (t ) d h( )exp( j )d (3.53) When the input is a weighted sum of signals with form exp(j 0 t), the output can be determined according to (3.52) and the linearity of the system.

28 Response to a Periodic Signal Let the input of a linear time-invariant continuous-time system be periodic. hen the output has the same period. he relation between the input and the output can be expressed as Y(k) 2 X(k)H k. (3.54) X(k) and Y(k) are the Fourier series coefficients of the input and the output, respectively. H() is the frequency response of the system. is the period. Proof. x(t) can be expressed as 2 x(t) X(k) exp j kt. (3.55) k According to (3.52) and the linearity of the system, we obtain

29 y(t) k and thus (3.54) is derived. 2 X(k)H Response to a General Signal k exp 2 j kt, (3.56) he I/O relation of a linear time-invariant continuous-time system can be expressed by the frequency response, i.e., Y()=X()H(), (3.57) where X() and Y() are the Fourier transforms of the input and the output, respectively, and H() is the frequency response. Proof. x(t) can be expressed as x(t) 2 X( )exp jt d. According to (3.52) and the linearity of the system, we obtain (3.58)

30 y(t) 2 X( )H( )exp jt d, (3.59) and thus (3.57) is derived. Also, (3.57) can be directly derived from the convolution property of the Fourier transform. Example. Let a linear time-invariant continuous-time system have the input x(t)=cos( 0 t) and the frequency response H()=exp( 2 ). Find the output. Example. A distortionless transmission system is described as Find the frequency response. y(t)=ax(tt 0 ). (3.60) Example. he phase of the frequency response is referred to as the phase response. he minus derivative of the phase response is called the group delay. Show that a distortionless transmission system has a constant group delay.

31 Example. An ideal low-pass filter has the frequency response Find the impulse response., 0 H( ). (3.6) 0, 0 Example. Find the responses of the above system to the following excitations: () x(t), t / 0 0, t / 0 over period 2 0, 2. 0 sin( 20t) ( 2) x(t). t Example. An ideal filter is best in frequency selectivity. However, it is noncausal, and its impulse response is oscillatory. hese defects

32 can be overcome by a non-ideal filter. he frequency response of a typical non-ideal low-pass filter is Find the impulse response. 0 H( ), (3.62) j Example. A band-pass filter can be implemented using a low-pass filter, as shown in figure 3.. Explain how it works. exp(j 0 t) exp(j 0 t) 0 x(t) e(t) Low-Pass Filter r(t) y(t) Figure 3.. Implementation of a Band-Pass Filter.

33 3.6. Linear Constant-Coefficient Differential Equations he zero-state response of a linear constant-coefficient differential equation can be found using the Fourier transform. Example. A causal, stable continuous-time system is given by dy(t) dt 2y(t) x(t), (3.63) where x(t)=e t u(t) and y(0 )=2. Find the zero-input response, the zero-state response, and the complete response. First, using the method in section 2.5, we can obtain the zero-input response y zi (t)=2e 2t. (3.64) hen, let us consider the zero-state response. he zero-state response satisfies (3.63), i.e.,

34 dyzs (t) dt 2y (t) x(t). (3.65) aking the Fourier transform of (3.65), we obtain jy From (3.66), we obtain Y zs zs zs ( ) 2Y zs ( ) he inverse Fourier transform of Y zs () is. j (3.66) ( ). (3.67) j 2 j y zs (t)=(e 2t +e t )u(t). (3.68) he complete response is the sum of the zero-input response and the zero-state response, i.e., y(t)= 2e 2t +(e 2t +e t )u(t). (3.69)

35 Example. A stable continuous-time system is given by dy(t) dt Find the impulse response. he zero-state response satisfies (3.70), i.e., dyzs (t) dt 2y(t) x(t). (3.70) 2y (t) x(t). (3.7) aking the Fourier transform of (3.7), we obtain Y zs ( ) X( ) zs. j 2 (3.72) hus, H( ). (3.73) j 2

36 he inverse Fourier transform of H() is h(t)=e 2t u(t). (3.74)