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UK EE 422G - 8-4A Difference Equation and Discrete-Time Systems

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EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-228-4A Difference Equation and Discrete-Time Systems Definitions of various discrete-time properties: Properties Definitions Linearity H[a⋅x1(nT)+ b⋅x2(nT)]= a⋅H[x1(nT)]+b⋅H[x2(nT)] Shift-Invariant H[x(nT-kT)] = y(nT-kT) Casual y(nT) depends on x(kT) for k≤n BIBO Stable x(nT) is bounded ⇒ y(nT) is bounded These are not new concepts – you have seen them in continuous-time system. In fact, one can treat discrete-time system as continuous-time system by using the continuous-time surrogates xs(t) and ys(t). On the other hands, a strong motivation to use discrete-time system is its ease of implementation using digital logic. Thus, we want to do all things “discrete”. In this section, we want to show: 1. The output of all Linear Shift-Invariant (LSI) discrete-time system can be computed by convolution with its IMPULSE RESPONSE. 2. Determine if a LSI system is causal and/or BIBO stable based on its impulse response. 3. Alternative representation of LSI system: difference equation 4. Classification and implementation of impulse responses: FIR and IIR LSI System Similar to the continuous-time relationship∫−=ττδτdtxtx )()()(, any discrete signal x(nT) can be written as ∑∞−∞=−=kkTnTkTxnTx )()()(δ Discrete-Time System H(⋅) x(nT) y(nT)EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-23Feed x(nT) to a LSI system H : []∑∞−∞=−==kkTnTkTxHnTxHnTy)]()([)()(δ ∑∞−∞=−=kkTnTHkTx )]([)(δ Linearity Define the impulse response )]([)( nTHnThδ=. As H is shift invariant, we have )]([)( kTnTHkTnTh−=−δ Thus, )()(ˆ)()()(nThnTxkTnThkTxnTyk∗=−=∑∞−∞= We can actually simplify the convolution expression a bit. First, assume x(kT) = 0 for k<0, we have ∑∞=−=0)()()(kkTnThkTxnTy What if the system is also causal? If so, y(nT) can only depend on x(kT) for k≤0. This implies that 00)(0)(<=⇔<=−nfornThnkforkTnTh For a casual LSI system, ∑=−=nkkTnThkTxnTy0)()()( Notice that the summation reduces from infinitely many terms to only n+1 terms. The output of any discrete-time LSI system can be written as the convolution between the input x(nT) and the impulse response h(nT).EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-24Let’s do one convolution by hand Note that the duration of the output y(nT) is longer than that of x(nT). In general, if x(nT) has N samples and h(nT) has K samples, y(nT) has N+K-1 samples. 0 2 2 1 1 x(kT) 0 2 1 h(kT) 0 y(0T)= Σkx(kT)h(0-kT) y(nT) 4 6 4 3 1 y(T)= Σkx(kT)h(T-kT) y(3T)= Σkx(kT)h(3T-kT) y(2T)= Σkx(kT)h(2T-kT) y(4T)= Σkx(kT)h(4T-kT) k k k k k k k nEE 422G Notes: Chapter 8 Instructor: Cheung Page 5-25Z-transform in action Time convolution is easy for finite duration impulse response. What if the impulse response is infinitely long? Answer: Use Z-transform We want to show for any causal LSI system: )()()()()()(0zXzHzYkTnThkTxnTyk=⇔−=∑∞= [ ])()()()()()()()()()()()()}()({))((0000 00 00zXzHzHzkTxzlThzkTxzlThkTxzkTnThkTxzkTnThkTxkTnThkTxZnTyZkkklklknkklklknlk nnnnkk====−=−=−=∑∑ ∑∑ ∑∑ ∑∑ ∑∑∞=−∞=−∞−=−=−−∞−=−=∞=∞=−∞=−∞=∞= Again, we call H(z) the TRANSFER FUNCTION of the discrete-time system. Example: )(41)( nTunTxn= )(31)( nTunThn= Find x(nT)*h(nT)EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-26Solution: First, compute the Z-transform X(z) and H(z) 14111)(41)(−−==znTuZzXn 13111)(31)(−−==znTuZzHn Then compute their product: )311)(411(1)()()(11 −−−−==zzzHzXzY Apply partial fraction : 111131144113)311)(411(1)(−−−−−+−−=−−=zzzzzY Then, )())31(4)41(3())(()(1nTuzYZnTynn+−==− DONE! Just like Laplace transform, we can also determine whether a system is BIBO stable based on the pole locations of the transfer function. Recall H(s) is BIBO stable if its ROC region includes the imaginary axis. And the imaginary axis in the s-plane maps to the unit circle in z-plane. Thus, we have H(z) is BIBO stable if and only if its ROC contains the unit-circle in the z-plane.EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-27 Types of Impulse Responses & Difference Equation Two main types of impulse responses: I. Finite Impulse Response (FIR) - h(nT) has finitely many non-zero values. Example H(z) = 1+2z-1+z-2 = (1+z-1)2 - Easy to see that if H(z) is FIR, there is no denominator so it does not have any finite poles ⇒ the corresponding system is always BIBO stable Express it in terms of the input x(nT) and output y(nT). Recall )()(2)()(21)()()(2121zXzzXzzXzYzzzHzXzY−−−−++=++== Taking the inverse Z-transform, we have )2()(2)()( TnTxTnTxnTxnTy−+−+= (1) 15.011)(−−=zzH BIBO Stable ))5.15.0(1)()5.15.0(1(1)(11 −−−−+−=zjzjzH Unstable n=0 1 2 1 h(nT)EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-28This is similar to the approach of recovering the differential equation from the transfer function in continuous-time. In discrete-time, (1) is called the Difference Equation. In general, the difference equation of any FIR system can be expressed as follows: ∑=−=NkkkTnTxanTy0)()( Why difference equation? Great for implementation – the followings are from Section 9-1, 9-2. Example: )2()(2)()( TnTxTnTxnTxnTy−+−+= Using simple delay, amplifier, and summer, we can implement this difference equation as: This structure is called the Direct Form. However, this is not the only type of implementation. Notice that the corresponding Z-transform of the difference equation is 2121)1(21)(−−−+=++= zzzzH Thus, the system can also be implemented


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