DIGITAL SIGNAL PROCESSING UNIT-I PART-A DEPT. / SEM.: CSE/VII. Define a causal system? AUC APR 09 The causal system generates the output depending upon present and past inputs only. A causal system is non anticipatory. 2. What are the properties linear system should satisfy? AUC NOV 08 A linear system should follow superposition principle. A linear system should satisfy, F[a x (t)+a 2 x 2 (t)] = [a y (t)+a 2 y 2 (t)] where Y (t) = f[x (t)] Y 2 (t) = f[x 2 (t)] 3. What is the criteria for the system to possess BIBO stability? AUC MAY 06 input. A system is said to be BIBO stable if it produces bounded output for every bounded 4. Define shift invariance? AUC NOV 07 If the system produces same shift in the output as that of input, then it is called shift invariance or time invariance system. i.e., F[x(t t )] = y(t t ) 5. Define (a) Random signal and (b) Deterministic signal? AUC NOV 09 Random signal: A random signal has some degree of uncertainty before it actually occurs. The random signal cannot be defined by mathematical expression. Deterministic signal: There is no uncertainly about the deterministic signal. It is completely represented by mathematical expression.
6. Classify the following signals as energy signals or power signals and find the normalized energy /power. Signals are defined over - <f < (a) cost+2 cos 2t (b)e 2 t.cost+2cosw2t AUC APR 0 This is a periodic signal. Hence it is power signal. Its power is given as, T 0 /2 P=/T 0 (cost+cos2t)dt =5/2 -T 0/2 e 2 t : This is non periodic signal. Hence it is energy signal. 7. Determine whether the following systems are time invariant or not. AUC NOV 07 y(t) = Sinx(t) y(t) = x(t) (a)y(t) = Sinx(t): y(t) = sinx(t) Let us determine the output of the system for delayed input x(t-t ).i.e., Y(t,t ) = f(t t )] = sin x (t t ) Here y(t,t ) represents output due to delayed input. Now delay the output y(t) by t. Hence we have to replace t by t-t in y(t) = sinx(t). i.e., y(t t ) = Sin x (t t ) On comparing the above equation with equation (a) we find that y(t t ) = y (t t ) Hence the system is time invariant. Any signal can be broken into a sum of two signals, one of which is odd equal to and other equal to.. Even part x e (t) = ½[x(t)+x(-t)] Odd part x 0 (t) = ½[x(t)-x(-t)] m 9. Is the discrete time system described by the equation y(n) = /2m+ x(n k) k= -m
Causal or noncausal? Why? AUC MAY 08 In the given equation when k = - m, y(n) = /2m+ x(n+m) Thus output at n th moment depends upon future input x(n+m). This makes the system noncausal. 0. Is the discrete time system described by the expression y(n) = nx(n) time invariant or not? Why? AUC NOV 06 Is the system described by following equation stable or not? Why? Y(n) = nx(n) The given discrete time system equation is (a) Y (n) = T [x (n)] = nx (n) When input x(n) is delayed by k samples the response is (b) y(n,k) = T[x(n-k)] = nx(n-k) Here observe that only input x(n) is delayed. The multiplier n is not part of the input. Hence it cannot be written as (n-k). Now let us shift or delay the output y(n) given by equation (a) by K samples. i.e., y(n,k) = (n-k)x(n-k) Here both n and x(n) in the equation y(n) = nx(n) will be shifted by k samples since they are part of the output sequence. It is clear from equation (b) and equation (c) that Y (n,k) y(n-k) Hence the system is shift variant. Is the system described by following equation stable or not? Why? AUC MAY 09 Y(t) = x( ) d t - This is unstable system, since it integrates function x(t) from -. Hence even though x(t) is bounded, the integration can be unbounded.
2. Define periodic and non-periodic signals. Give an example in each case? AUC MAY 08 A periodic signal repeats after fixed period. But nonperiodic signal never repeats. Periodic signal : x(t) = Sin t Non periodic signal : x(t) = e 2t A discrete time signal is periodic if its frequency can be expressed as ratio of two integers. i.e., Frequency f 0 = K/N Here K and N are integer and N is the period of discrete time signal. 3. Distinguish between energy and power signals. AUC NOV 0 For the power signal, normalized average power is finite and nonzero. Energy of the power signal is infinite. For the energy signal, total normalized energy is finite and nonzero. Power of the energy signal is zero. More details are given in Table Power signal. The normalized average power is finite and nonzero. 2. Practical periodic signals are power signals 3. These signals can exist over infinite time. 4. Energy of the power signal is infinite over infinite time. Energy Signal Total normalized energy is finite and nonzero. Non periodic signals are energy signals. These signals are time limited. Power of the energy signal is zero over infinite time. 4. Is the discrete time system described by the equation y(n) = x(-n) causal or noncausal? Why? AUC MAY 04 Here y(n) = x(-n). If n = -2 then, Y(-2) = x(2) Thus the output depends upon future inputs. Hence system is noncausal.
5. Is the system described by the equation y(t) = x(2t) time invariant or not? Why? AUC NOV Output for delayed input becomes, Y(t,t ) = x(2t t ) Delayed output will be, Y(t-t ) = x[2(t t )] Since y(t,t ) y (t t ) the system is shift variant. 6. Check whether the given system is causal and stable? AUC APR 06 Y(n) = 3x(n 2) +3x(n+2) Since y(n) depends upon x(n+2), this system is noncausal. As long as x(n-2) and x(n+2) are bounded, the output y(n) will be bounded. Hence this system is stable. 7. When the discrete signal is said to be even? AUC NOV 05 A discrete time signal is said to be even when, x(-n) = x(n) For example cos n is an even signal. 8. Represent the signal u r (n) =n for 0. AUC NOV 06 n for n < 0 This signal is a ramp signal. 9. What do you mean by energy of a signal? AUC APR 08 E = x(t) 2 dt for continuous signal. - E = x(n) 2 dt for discrete signal. N= -
20. Check causality of the system given by, AUC APR 09 y(n) = x(n n 0 ) If n 0 0, then output y(n) de[ends upon present or past input. Hence the system is causal. n 0 <0, the system becomes noncausal. PART-B.Explain the Classification of CT & DT signals. AUC APR 05, MAY, NOV 06 Signals are classified into 2 categories I ) Continuous time signals II) Discrete time signals I) Continuous time signals: The signal that is specified for every value of time t is called Continuous time signal and it is denoted by t. II) Discrete time signals Discrete signals are represented as a sequence of numbers and it is denoted by n Classification of CT and DT Signals:. Deterministic and random signals 2. Periodic and periodic signals 3. Energy and power signals 4. Even and odd signals
. Deterministic and random signals: A Signal is said to be deterministic signal if it is expressed by mathematical expressions. Ex: x(t)=a Sint A Signal is said to be random signal which cannot be easily expressed or determined by mathematical expressions. 2. Periodic and Aperiodic signals A signal is said to be periodic signal which maintains the regular intevels is called periodic signals A signal is said to be Aperiodic signal which is not in the regular intervals 3. Energy and power signals A signal is said to be Energy signal if the total energy of the signal is finite and non zero. A signal is said to be Power signal if the total power of the signal is finite and non zero Expression for energy and power signals- CT Signal T E=Lt T- -T T x (t) 2 dt P= Lt /2T x (t) 2 dt T- -T
Expression for energy and power signals- DT Signal E= x(n) 2 n=- N P= Lt /(2N+) x(n) 2 n=- n= -N 4. Even and Odd signals: A signal is said to be even signal it satisfies the following relationship.. Even signals are also called as Symmetric signals. x(-t)= x(t) for all t; x(-n)= x(n) for all n A signal is said to be odd signal it satisfies the following relationship. Odd signals are also called as Anti Symmetric signals. x(-t)= -x(t) for all t; x(-n)= -x(n) for all n Even and odd part of the signals: Even part of the signals can be denoted as X e (t) and Odd part of the signal can be denoted as X o (t). X e (t) = ½ [x(t)+x(-t)] X o (t) =½ [x(t) - x(-t)] For CT Signals X e (n) = ½ [x(n)+x(-n)] X o (t) =½ [x(n) - x(-n)] For DT Signals
2. Explain Classification of CT & DT Systems AUC APR 07, MAY 08, NOV 09. Linear and Non linear systems 2. Causal and non causal systems 3. Static and dynamic systems 4. Time invariant and time variant systems 5. stable and unstable systems. Linear and Non linear systems A System is said to be linear system ii satisfies the superposition principle. Consider the systems y (t), y 2 (t). y (t) = F{ x (t)} y 2 (t) = F{ x 2 (t)} y 3 (t)= a y (t)+a 2 y 2 (t) y 3 (t)= F{a x (t)+a 2 x 2 (t)} y 3 (t)= y 3 (t) ------ Linear system y 3 (t) y 3 (t) ------ Non Linear system 2. Causal and Non causal system: A System is said to be causal system if the output of the system depends on the present and past outputs Ex: x(t). x(t-a) A system is said to be non causal if the output of the system depends on future inputs also. Ex: x(t 2 ), x(t+a)
3. Static and dynamic systems: A System is said to static if the output depends on present input only. Ex: x(t)cos t A System is said to be dynamic if the output depends on past and future inputs 4. Time invariant and time variant system: A System is said to be time invariant if the time shift in the input signal results in corresponding time shift in the output Let y(t)=f{x(t)}, if x(t) is delay by time t, then the output will also be with same delay. The time invariant system satisfies the following relationship F{x(t-t )}=y(t-t ) y(t, t ) = y(t-t ) ------ Time invariant system y(t, t ) y(t-t ) ------ Time variant system 5. Stable and unstable systems: When the system produces bounded output for the bounded input then the system is called stable system and also called BIBO stable systems. When the system produces unbounded output for the bounded input it is called unstable systems.
3. Sketch the following type of signals. AUC MAY 08, APR 06 i) u (t-2) ii) 2u (t-2) iii) -3u (t-2) iv) u(-t+) Solution: (i) u (t-2) u(t) a t u (t-2) 2 t ii) 2u (t-2) 2u(t-2) 2 2 t
iii) -3u (t-2) u (t-2) 2 t 2 t -3-3u (t-2) iv) u(-t+) u(t) t u(-t)
-t u(-t+) -t 0 t 4. Sketch the signal x (t), Plot x(t+2), x(-t+) AUC MAY 09, MAY 2, NOV 04 0 2 t
X (t+2) -t -2-0 X (-t+) X (-t) -t -2-0 X (-t+) -t - 0 t
5. Determine whether the systems are Static/Dynamic, causal/non causal, Linear/Non linear, time invariant/variant systems. AUC MAY 05, MAY 2, a) y(t)=0x(t)+5 b) y(t)= x(t) c) y(n)=cos[x(n)] a) y(t) =0x (t)+5 I) Linear/ Non linear system: y(t)=0x(t)+5 y (t)=0x (t)+5 y 2 (t)=0x 2 (t)+5 y 3 (t)=a y (t)+a 2 y 2 (t) y 3 (t)=a {0x (t)+5}+a 2 {0x 2 (t)+5} y 3 (t)=f{a x (t)+a 2 x 2 (t)} y 3 (t)=0{ a x (t)+a 2 x 2 (t)}+5 y 3 (t) y3 (t) The given system is Non linear II) Static/dynamic: Static The output of the system is depends on the present input. Hence the system is III) Causal /Non causal system: The output of the system is depends on the present input. Hence the system is Causal IV) Time invariant/variant system: y (t,t )= 0 x(t-t )+5 y (t-t )= 0 x(t-t )+5 y (t,t ) = y(t-t ) The given system is time invariant.
b) y(t)= x(t) I) Linear/ Non linear system: y(t)= x(t) y (t)= x (t) y 2 (t)= x 2 (t) y 3 (t)=a y (t)+a 2 y 2 (t) y 3 (t)=a x (t) +a 2 x 2 (t) y 3 (t)=f{a x (t)+a 2 x 2 (t)} y 3 (t)= a x (t)+a 2 x 2 (t) y 3 (t) y3 (t) The given system is Non linear II) Static/dynamic: Static The output of the system is depends on the present input. Hence the system is III) Causal /Non causal system: The output of the system is depends on the present input. Hence the system is Causal IV) Time invariant/variant system: y (t,t )= x(t-t ) y (t-t )= x(t-t ) y (t,t ) = y(t-t ) The given system is time invariant C) y(n)=cos[x(n)] I) Linear/ Non linear system: y(n)=cos[x(n)] y (n)=cos[x (n)] y 2 (n)=cos[x 2 (n)] y 3 (n)=a y (n)+a 2 y 2 (n)
y 3 (n) =a {cos[x (n)]}+a 2 {cos[x 2 (n)]} y 3 (n) =F {a x (n)+a 2 x 2 (n)} y 3 (n) =cos {a x (n)+a 2 x 2 (n)} y 3 (n) y3 (n) The given system is Non linear II) Static/dynamic: Static The output of the system is depends on the present input. Hence the system is III) Causal /Non causal system: The output of the system is depends on the present input. Hence the system is Causal IV) Time invariant/variant system: y(n)=cos[x(n)] y(n,n )=cos[x(n-n )] y (n-n )=cos[x(n-n )] y (n,n )= y(n-n ) The given system is time invariant 6. Find whether the signals are periodic or aperiodic. If periodic determine its time period. AUC MAY 06 a) X(t)=cos(6л t) b) cos(0.3n) a) X (t) =cos (6л t) 2л ft = 6л t So, f=3 The given signal is periodic signal b) cos (0.3n) So, the time period T=/3 2л fn = 0.3n So, f=0.3/2л
The given signal is Aperiodic 7. Find whether the signals are periodic or aperiodic. If periodic determine its time period. AUC NOV 08 a) cost+sin3t b) cos (n/8) cos (nл/8) a) cost+sin3t 2л ft = t 2л f2t =3 t f=/2л, the given signal is aperiodic. f2=3/2л, the given signal is aperiodic. The entire signal is aperiodic b) cos (n/8) cos (nл/8) 2л fn = n/8 f=/6л, the signal is aperiodic 2л f2n = n л /8 f2=/6, T=6, The signal is periodic The entire signal is aperiodic 8. Sketch the even and odd components of the signals. AUC NOV 07 X (t) 2 0 3 t Solution: Even Signal X (t) X(-t) 2 2 0 3 t -t -3 0
X(t)+X(-t) 2 -t -3 0 3 t X e (t) = ½ [x(t)+x(-t)] ½(X(t)+X(-t)) -t -3 0 3 t Odd Signal -x(t) X (t) -t -3 2-2 0 3 t
-t -3-2 X(t)-X(-t) 0 3 t -2 -t -3 - ½(X(t)-X(-t)) 0 3 t - 9. Find the Even and odd components of the signals AUC MAY 08 3 0 2 t X(-t) 3 -t -2 0
3 X(t)+x(-t) -t -2 0 2 t.5 ½(X(t)+x(-t)) -t -2 0 2 t 0.Explain-basic properties of signals. AUC NOV. Time shifting property 2. Time reversal property 3. Amplitude scaling 4. Time scaling 5. Signal addition and multiplication. Time shifting property For CT Signals Consider the signal x (t). The time shifting of the signal x(t) may delay or advance. Let x (t) =u (t) - Unit step function
Unit step function may be defined as, u(t)=, t 0 0, t<0 u(t) (t) t U(t) a t u (t-a)=, t a 0, t<0 Consider the same signal u(t), time advanced at T seconds should noted as u(t+t)
u(t+t) -t 0 t u (t+t)=, t<-t 0, t<0 For DT Signals: Consider the signal x(n)= u(n) u(n) =, n 0 0, n<0 u(n) 0 2 3 4 5 6 n
u(n-2) 0 2 3 4 5 6 n u(n+2) -2-0 2 3 4 5 n 2. Time reversal property: Time reversal property of the signal x (t) is obtained by holding the signal at t=0. So it should be x (-t). Let x (t) =u (t) For CT Signals:
t u(-t) -t For DT Signals:
u(n) n u(-n) -n 3. Ampitude scaling: The signal x (t) is multiplied by any constant a. This is defined as amplitude scaling Consider the signal y(t)= a x(t). Let x(t)= u(t) u (t) =, t 0 0, t<0 Let y(t)= a u(t) and Let a=4, So y(t)= 4 u(t)
4 t t 4. Time scaling: X (at)= x{t/a) Let us consider the signal x (t). Here the time t can be multiplied by scaling factor a which is equivalent to x (at) which is defined as time scaling X(t) X(2t) 0 t 0 /2 t X (t/2) 0 2 t
5. Signal addition and Multiplication: The sum of two signals can be obtained by adding their values at every instant similarly the subtraction of two signals can be obtained by subtracting their values at every instant.