IntroductionLecture 1: Basic Ideas & Terminology DIGITAL CONTROL SYSTEM 자동제어 PROF. KALYANA C. VELUVOLU (별루볼루 교수) School of Electronics Engineering (전자공학부 IT 대학)

WEEK 4

Z-TRANSFORM AND PLANE TRANSFORMATIONS Topics: Z transform properties Inverse z transform

Z TRANSFORM PROPERTIES 1. Addition and Subtraction: Given 𝐸 𝑎 z =Z[ 𝑒 𝑎 ∗ t ]and 𝐸 𝑏 (𝑧)=Z 𝑒 𝑏 ∗ t , then Z 𝑒 𝑎 ∗ t + 𝑒 𝑏 ∗ t = 𝐸 𝑎 z + 𝐸 𝑏 (𝑧) 2. Multiplication by a constant: Given 𝐸 𝑧 =𝑍 𝑒 ∗ 𝑡 , and A=constant, then 𝑍 𝐴 𝑒 ∗ 𝑡 =𝐴𝑍 𝑒 ∗ 𝑡 =𝐴𝐸(𝑍) 3. Real translation (shifting): Given 𝐸 𝑍 =𝑍[ 𝑒 ∗ 𝑡 ], then shifting to the right (delay) gives 𝑍 𝑒 ∗ 𝑡−𝑝𝑇 = 𝑧 −𝑝 𝐸 𝑧 And shifting to the left (advance) gives 𝑍[e^∗(𝑡 + 𝑝𝑇)]= 𝑧 𝑝 𝐸 𝑧 − 𝑖=0 𝑝−1 𝑒(𝑖𝑇) 𝑧 𝑝−𝑖 Dr. Kalyana Veluvolu

4. Complex translation: Given 𝐸 𝑧 =𝑍 𝑒 ∗ 𝑡 and 𝐴=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑙𝑒𝑡 𝑒 𝐴 ∗ 𝑡 = 𝑒 ±𝐴𝑇 𝑒 𝑡 . 𝑡ℎ𝑒𝑛 𝑍 𝑒 𝐴 ∗ 𝑡 =𝑍 𝐸 ∗ 𝑠±𝐴 =[E(z)] with z replaced by ze^(±AT) =𝐸(𝑧 𝑒 ±𝐴𝑇 )= 𝐸 𝐴 (z) Dr. Kalyana Veluvolu

𝑐 𝑘𝑇 =−2𝑐 𝑘−1 +𝑐 𝑘−2 𝑇 +𝑒 𝑘𝑇 −0.5𝑒 𝑘−2 𝑇 Example: For the linear difference equation 𝑐 𝑘𝑇 =−2𝑐 𝑘−1 +𝑐 𝑘−2 𝑇 +𝑒 𝑘𝑇 −0.5𝑒 𝑘−2 𝑇 Property 3 can be used to determine 𝐺(𝑧) of the system, where 𝐺(𝑧) is the transfer function of the system. 𝐶 𝑧 =𝑍 𝑐 𝑘𝑇 =−2 𝑧 −1 𝐶 𝑧 + 𝑧 −2 𝐶 𝑧 +𝐸 𝑧 −0.5 𝑧 −2 𝐸 𝑧 1+2 𝑧 −1 + 𝑧 −2 𝐶 𝑧 = 1−0.5 𝑧 −2 𝐸 𝑧 𝐺 𝑧 = 𝐶(𝑧) 𝐸(𝑧) = 1−0.5 𝑧 −2 1+2 𝑧 −1 + 𝑧 −2 Dr. Kalyana Veluvolu

Example : To obtain Z transform of 𝑒 −𝐴𝑡 𝑐𝑜𝑠𝜔𝑡, first consider only 𝑒(𝑡)=cos𝜔𝑡, 𝐸 𝑧 =𝑍 𝑒 ∗ 𝑡 = 𝑧(𝑧−𝑐𝑜𝑠𝜔𝑇) 𝑧 2 −2𝑧 𝑐𝑜𝑠𝜔𝑇 +1 Then, replacing z by z 𝑒 +𝐴𝑡 yields Z[ 𝑒 𝐴 ∗ (t)]= z 𝑒 +𝐴𝑇 (z 𝑒 +𝐴𝑇 −𝑐𝑜𝑠𝜔𝑇) z 𝑒 +2𝐴𝑇 −2z 𝑒 +𝐴𝑇 𝑐𝑜𝑠𝜔𝑇 +1 = 𝑧 2 −z 𝑒 −𝐴𝑇 𝑐𝑜𝑠𝜔𝑇 𝑧 2 −2𝑧 𝑒 −𝐴𝑇 𝑐𝑜𝑠𝜔𝑇 + 𝑒 −2𝐴𝑇 Dr. Kalyana Veluvolu

𝐸 𝑧 =𝑍[ 𝑒 ∗ (t)= 𝑧(𝑧−𝑐𝑜𝑠𝜔𝑇) 𝑧 2 −2𝑧 𝑐𝑜𝑠𝜔𝑇 +1 Initial-Value Theorem: Given 𝐸 𝑧 =𝑧 𝑒 ∗ 𝑡 = 𝑘=0 ∞ 𝑒(𝑘𝑇) 𝑧 −𝑘 and supposing lim 𝑡→0 𝑘→0 𝑒 ∗ (𝑡)= lim 𝑛→∞ 𝐸(𝑧) Example 2.3: The initial value of the function of 𝐸 𝑧 =𝑍[ 𝑒 ∗ (t)= 𝑧(𝑧−𝑐𝑜𝑠𝜔𝑇) 𝑧 2 −2𝑧 𝑐𝑜𝑠𝜔𝑇 +1 Is evaluated by lim 𝑡=0 𝑒 ∗ (𝑡)= lim 𝑧→∞ 𝐸 𝑧 =1 Dr. Kalyana Veluvolu

Final- Value theorem: Given 𝐸 𝑧 =𝑍 𝑒 ∗ 𝑡 and that the function (1− 𝑧 −1 )E(z) does not have any poles outside or on the UC in the plane, then lim 𝑡→∞ 𝑒 𝑡 = lim 𝑡→∞ 𝑒 𝑘𝑇 = lim 𝑧→1 1− 𝑧 −1 𝐸(𝑧) Example 2.4: The final value of the function 𝐸(𝑧)= 2.4𝑧 (𝑧−0.5)(𝑧+0.2)(𝑧−1) Is evaluated by lim 𝑡→∞ 𝑒 ∗ 𝑡 = lim 𝑧→1 [(1− 𝑧 −1 𝐸(𝑧) = 2.4𝑧 (𝑧−0.5)(𝑧+0.2) =4 Dr. Kalyana Veluvolu

Real Convolution: Given that 𝐸 1 (z)=𝑍[ 𝑒 1 ∗ 𝑡 ] and 𝐸 2 𝑧 =𝑍[ 𝑒 2 ∗ 𝑡 ], then 𝐸 1 𝑧 𝐸 2 𝑧 =𝑍 𝑖=0 𝑘 𝑒 1 (𝑖𝑇) 𝑒 2 (𝑘𝑇−𝑖𝑇) =𝑍 𝑖=0 𝑘 𝑒 1 (𝑘𝑇−𝑖𝑇) 𝑒 2 (𝑖𝑇) = 𝑘=0 ∞ 𝑖=0 𝑘 𝑒 1 (𝑖𝑇) 𝑒 2 (𝑘𝑇−𝑖𝑇) 𝑧 −𝑘 Note that 𝑒 1 𝑘𝑇 𝑒 2 𝑘𝑇 = 𝑍 −1 [ 𝐸 1 𝑧 𝐸 2 𝑧 ]= 𝑖=0 𝑘 𝑒 1 𝑖𝑇 𝑒 2 𝑘𝑇−𝑖𝑇 Dr. Kalyana Veluvolu

Differentiation: Given that 𝐸 𝐴,𝑧 =𝑍 𝑒 ∗ 𝐴,𝑡 , where A is a constant or an independent variable, and 𝑓(𝐴,𝑡)= 𝜕 𝑒 𝑚 (𝐴,𝑡) 𝜕 𝐴 𝑚 Then 𝐹(𝐴,𝑧)=𝑍[𝑓 𝐴,𝑡 =𝑍 𝜕 𝑚 [ 𝑒 ∗ (𝐴,𝑡) 𝜕 𝐴 𝑚 = −1 𝑚 𝜕 𝑚 𝐸(𝐴,𝑧) 𝜕 𝐴 𝑚 , m>0 Note: For certain functions 𝑓(𝐴,𝑡), it is easy to obtain the z transform of the mth partial derivative of e( A, t), with respect to A. Dr. Kalyana Veluvolu

Example : Given 𝑓(𝐴,𝑡)= 𝑡 2 𝑒 −𝐴𝑡 , where 𝑒 𝐴,𝑡 = 𝑒 −𝐴𝑡 and 𝑓 𝐴,𝑡 = 𝜕 𝑒 2 (𝐴,𝑡) 𝜕 𝐴 2 and 𝐸 𝐴,𝑧 =𝑍 𝑒 ∗ 𝐴,𝑡 = 𝑧 𝑧− 𝑒 −𝐴𝑇 Then by property 8, we have Type equation here. 𝐹 𝐴,𝑧 =(−1 ) 2 𝜕 2 𝐸(𝐴,𝑧) 𝜕 𝐴 2 = 𝑇 2 𝑒 −𝐴𝑇 𝑧(𝑧+ 𝑒 −𝐴𝑇 ) (𝑧− 𝑒 −𝐴𝑇 ) 3 Dr. Kalyana Veluvolu

THE INVERSE THEOREM Partial-Fraction Method Consider F(z) of the form 𝐸 𝑧 = 𝐾( 𝑧 𝑤 + 𝑐 𝑤 𝑧 𝑤−1 +…+ 𝑐 1 𝑧+ 𝑐 0 𝑧 𝑛 + 𝑑 𝑛−1 𝑧 𝑛−1 +…+ 𝑑 1 𝑧+ 𝑑 0 = 𝐾𝑁(𝑧) 𝐷(𝑧) where 𝑐 0 ≠0. Note that E(z) has w zeros and n poles. If w>n, divide N(z) by D(z) until the order of the remainder polynomial is at least one degree less than D(z). This will yield 𝐸 𝑧 =𝐾[ 𝑔 𝑟 𝑧 𝑟 + 𝑔 𝑟−1 𝑧 𝑟−1 +…+ 𝑔 1 𝑧+ 𝑔 0 + 𝐸 ′ 𝑧 ] where 𝐸 ′ 𝑧 = 𝑘 𝑛−1 𝑧 𝑛−1 +…+ 𝑘 1 𝑧+ 𝑘 0 𝑧 𝑛 + 𝑑 𝑛−1 𝑧 𝑛−1 +…+ 𝑑 1 𝑧+ 𝑑 0 = 𝑁 ′ (𝑧) 𝐷(𝑧) Dr. Kalyana Veluvolu

Case 1: 𝐸 ′ 𝑧 has zeros at the origin ( 𝑘 0 =0, 𝑐 0 ≠0) Step 1. Obtain 𝐸 ′ 𝑧 𝑧 Step 2. Perform partial-fraction expansion Step 3. Multiply the resulting equation of step 2 by z. Step 4. Obtain 𝑒 ′∗ 𝑡 by taking the inverse transform of the result in step 3. Step 5. compute 𝑒 ∗ 𝑡 if n>w. The first three steps ensure that each function in the expansion has zero at the origin. Dr. Kalyana Veluvolu

Example: Obtain inverse z-transform for 𝐸 𝑧 = 10( 𝑧 3 − 𝑧 2 +3𝑧−1) (𝑧−1)( 𝑧 2 −𝑧+1) Since w=n, it is necessary to first divide N(z) by D(z) to yield 𝐸 𝑧 =10 1+ 𝑧(𝑧+1) (𝑧−1)( 𝑧 2 −𝑧+1) As 𝐸 ′ 𝑧 has one zero at the origin, we follow Case 1 procedure: Step1 𝐸 ′ 𝑧 𝑧 = 𝑧+1 (𝑧−1)( 𝑧 2 −𝑧+1) = 𝑧+1 (𝑧−1)(𝑧− 𝑎 1 )(𝑧− 𝑎 2 ) 𝑎 1 = 0.5+j0.5 3 = 𝑎 1 < 𝜋 3 =exp( 𝑏 1 𝑇)= 𝑒 0 𝑒 𝑗𝜋 3 𝑎 2 =0.5−j0.5 3 = 𝑎 2 <− 𝜋 3 = exp( 𝑏 2 𝑇)= 𝑒 0 𝑒 −𝑗𝜋 3 where 𝑎 1 = 𝑎 2 =1, 𝑏 1 =0+j𝜋3, and 𝑏 2 =0−𝑗𝜋3. Dr. Kalyana Veluvolu

Step 2 𝐸 ′ (𝑧) 𝑧 = 2 (𝑧−1) − 1 (𝑧− 𝑒 𝑗𝜋 3 ) − 1 (𝑧− 𝑒 − 𝑗𝜋 3 ) =2 𝑧 (𝑧−1) − 𝑧(𝑧− cos 𝜔𝑇) 𝑧 2 − 2𝑐𝑜𝑠𝜔𝑇 +1 𝜔𝑇= 𝜋 3 Step 4. Using the Transform Table, 𝑒 ′∗ 𝑡 = 𝑘=0 ∞ (2− 𝑒 𝑗( 𝜋 3)𝑘 + 𝑒 −𝑗( 𝜋 3)𝑘 )𝛿(𝑡−𝑘𝑇) = 𝑘=0 ∞ (2−2𝑐𝑜𝑠𝑘𝜔𝑇) 𝛿(𝑡−𝑘𝑇) or 𝑒(𝑘𝑇) ′ =2−2𝑐𝑜𝑠𝑘𝜔𝑇 for k≥0 Dr. Kalyana Veluvolu

Step 5. Thus, 𝑒 ∗ 𝑡 =10 𝑍 −1 1 +10 𝑒 ∗ (𝑡 ) ′ =10𝛿 𝑡 + 𝑘=0 ∞ (20−20𝑐𝑜𝑠𝑘𝜔𝑇)𝛿(𝑡−𝑘𝑇) or e(kT)=10𝛿 𝑡 +20−20𝑐𝑜𝑠𝑘𝜔𝑇 for k≥0 Dr. Kalyana Veluvolu

Case 2: 𝐸 ′ (z) has zeros at the origin ( 𝑘 0 =0, 𝑐 0 =0) Step 1 Case 2: 𝐸 ′ (z) has zeros at the origin ( 𝑘 0 =0, 𝑐 0 =0) Step 1. Expand 𝐸 ′ (z). Step 2. Multiply the resulting equation of step 1 by z to obtain 𝐸 𝑑 ′ 𝑧 =𝑧 𝐸 ′ 𝑧 . Step 3. Obtain 𝑒 𝑑 ′∗ 𝑡 by taking the inverse transformation of the result in step 2 Step 4. Since 𝑒 𝑑 ′∗ 𝑡 = 𝑍 −1 [ 𝐸 𝑑 ′ (z)]= 𝑍 −1 [ 𝑧𝐸 ′ (𝑧)] Property 3 implies 𝑒 ∗ (𝑡 ) ′ = 𝑍 −1 𝐸 ′ 𝑧 = 𝑍 −1 [ 𝑧 −1 𝐸 𝑑 ′ 𝑧 ] = 𝑘=1 ∞ 𝑒 𝑑 ′ [ 𝑘−1 𝑇]𝛿(𝑡−𝑘𝑇) Step 5. Compute 𝑒 ∗ 𝑡 𝑖𝑓 𝑤>𝑛 Dr. Kalyana Veluvolu

Example: Obtain inverse z transform for 𝐸 𝑧 = −10(11 𝑧 2 −15𝑧+6) (𝑧−2)(𝑧−1 ) 2 =10 𝐸 ′ (𝑧) (K=10) Step 1 𝐸 ′ 𝑧 = 𝐴 11 (𝑧−1) + 𝐴 12 (𝑧−1) 2 + 𝐴 2 (𝑧−2) → 𝐸 ′ 𝑧 =− 9 (𝑧−1) + 2 (𝑧−1) 2 + 20 (𝑧−2) Step 2 𝐸 𝑑 ′ 𝑧 =𝑧 𝐸 ′ 𝑧 =− 9𝑧 (𝑧−1) + 2𝑧 (𝑧−1) 2 + 20𝑧 (𝑧−2) Dr. Kalyana Veluvolu

Step 3 𝑒 𝑑 ′ 𝑘𝑇 =−2𝑘−9+20 𝑒 0.693𝑘 for k>0 Step 4 𝑒 ′ 𝑘𝑇 =−2 𝑘−1 −9+20 𝑒 0.693(𝑘−1) for k>0 Step 5 e(kT)=K 𝑒 ′ (kT)=20(k-1)+90-200 𝑒 0.693(𝑘−1) for k>0 To obtain 𝑒 ′ 𝑘𝑇 , which is 𝑒 𝑑 ′ (kT) delayed by one period, replace k on the right-hand side of the equation for 𝑒 𝑑 ′ 𝑘𝑇 by k-1 which yields Dr. Kalyana Veluvolu

Power Series Method (Direct Division Method) The method is to obtain the desired values 𝑒(𝑘𝑇) by dividing 𝑁 𝑧 by 𝐷(𝑧) to obtain a power series in terms of powers of 𝑧 −1 . This method results in an open form of E 𝑧 and in turn for 𝑒 ∗ 𝑡 . Dividing N(z) by the D(z) for 𝐸 𝑧 = 𝐾( 𝑧 𝑤 + 𝑐 𝑤−1 𝑧 𝑤−1 +…+ 𝑐 1 𝑧+ 𝑐 0 ) 𝑧 𝑛 + 𝑑 𝑛−1 𝑧 𝑛−1 +…+ 𝑑 1 𝑧+ 𝑑 0 = 𝐾𝑁(𝑧) 𝐷(𝑧) yields 𝐸 𝑧 =𝐾[ 𝐵 𝑤−𝑛 𝑧 𝑤−𝑛 + 𝐵 𝑤−𝑛−1 𝑧 𝑤−𝑛−1 +…] Recall 𝐸 𝑧 = 𝑘=0 ∞ 𝑒(𝑘𝑇) 𝑧 −𝑘 Dr. Kalyana Veluvolu

We have K 𝐵 𝑤−𝑛 =𝑒 𝑛−𝑤 𝑇 K 𝐵 𝑤−𝑛−1 =𝑒 𝑛−𝑤+1 𝑇 which are the values of e(t) at the sampling instants 𝑘𝑇= 𝑛−𝑤 𝑇, 𝑛+1− 𝑤 ′ … Dr. Kalyana Veluvolu

Example : Repeat previous Example by using the power series method and compare the results. 𝐸 𝑧 = −10(11 𝑧 2 −15𝑧+6) (𝑧−2) (𝑧−1) 2 =10 𝑁(𝑧) 𝐷(𝑧) 𝑁 𝑧 =11 𝑧 2 −15𝑧+6 11 𝑧 2 −44𝑧+55−22 𝑧 −1 29𝑧−49+22 𝑧 −1 29𝑧−116+145 𝑧 −1 −58 𝑧 −2 67−123 𝑧 −1 +58 𝑧 −2 67−268 𝑧 −1 … 𝐷 𝑧 = 𝑧 3 −4 𝑧 2 +5𝑧−2 𝐸 𝑧 =11 𝑧 −1 +29 𝑧 −2 +67 𝑧 −3 +145 𝑧 −4 +… →𝑒 0 =0, 𝑒 𝑇 =11, 𝑒 2𝑇 =29, 𝑒 3𝑇 =67… Dr. Kalyana Veluvolu

Remarks The use of an ideal sampler in the analysis, for this mathematical model to be close to correct, it is necessary for the sampling duration 𝛾 𝑡𝑜 𝑏𝑒 very small by comparison with the smallest time constant of the E(s) and T. The inverse z-transform, 𝑍 −1 [E(z)], may not yield a unique e(t) function. Dr. Kalyana Veluvolu