dis04B_sol

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EECS 16B Designing Information Systems and Devices II UC Berkeley Fall 2023 Discussion 4B 1. Transfer Function Practice Transfer functions take an input phasor and “transform” it into an output phasor. Most of the work we will do with transfer functions is analyzing how it will “respond” to a specific kind of input. We will also design our own transfer functions using common circuit components such as resistors, inductors, and capacitors to achieve some specified behavior. A block diagram of a transfer function is represented below. In this discussion, we will learn how to derive H ( j ω ) from a given circuit, and we will analyze how it behaves for certain values of ω . e V in ( j ω ) H ( j ω ) e V out ( j ω ) Figure 1: Transfer Function Block Diagram Recall that Z L = j ω L and Z C = 1 j ω C . For large ω , | Z L | = ω L becomes large (and becomes small for small ω ). On the other hand, for large ω , | Z C | = 1 ω C becomes small (and becomes large for small ω ). In this problem, you’ll be deriving some transfer functions. For each circuit: • Determine the transfer function H ( j ω ) = e V out ( j ω ) e V in ( j ω ) . • How does | H ( j ω ) | respond as ω → 0 (low frequencies) and as ω → ª§¦ª (high frequencies) ? • Is the circuit a high-pass filter, low-pass filter, or band-pass filter ? • For parts (a) and (b) , find the cutoff frequency ω c , which is the frequency such that | H ( j ω c ) | = | H ( j ω ) | max √ 2 (1) 1

EECS 16B Discussion 4B 2023-09-20 18:03:56-07:00 (a) RC circuit ( R = 1 k \YX , C = 1 µF): + − v in ( t ) C R + − v out ( t ) (a) Circuit in “time domain” − + e V in ( j ω ) Z C ( j ω ) Z R ( j ω ) + − e V out ( j ω ) (b) Circuit in “phasor domain” Solution: We’ll use the voltage divider formula to find e V out ( j ω ) : e V out ( j ω ) = Z R Z R + Z C e V in ( j ω ) (2) Recalling the expression for the impendances, we note that for the resistor Z R = R , and for the capacitor Z C = 1 j ω C . Plugging in the impedances gives H ( j ω ) = e V out ( j ω ) e V in ( j ω ) = R R + 1 j ω C = j ω RC 1 + j ω RC (3) At low frequencies, we have lim ω → 0 | H ( j ω ) | = lim ω → 0 ω RC √ 1 + ω 2 R 2 C 2 = 0 (4) At high frequencies, we have lim ω → ª§¦ª | H ( j ω ) | = lim ω → ª§¦ª ω RC √ 1 + ω 2 R 2 C 2 (5) = lim ω → ª§¦ª ω RC √ ω 2 R 2 C 2 (6) = 1 (7) So this circuit is a high-pass filter. For this transfer function, | H ( j ω ) | max = 1. Thus, to find the cutoff frequency ω c , we need to find when | H ( j ω c ) | = 1 √ 2 . | H ( j ω c ) | = 1 √ 2 (8) ω RC p 1 + ω 2 c R 2 C 2 = 1 √ 2 (9) 1 + ω 2 c R 2 C 2 = 2 ω 2 R 2 C 2 (10) ω c = 1 RC (11) = 1 ( 10 3 )( 10 − 6 ) = 10 3 rad s (12) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 2

EECS 16B Discussion 4B 2023-09-20 18:03:56-07:00 Notice that this can be observed from the transfer function itself by writing it in the following form: j ω RC 1 + j ω RC = j ω 1 RC 1 + j ω 1 RC = j ω ω c 1 + j ω ω c (13) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 3

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EECS 16B Discussion 4B 2023-09-20 18:03:56-07:00 (b) LR circuit ( L = 5 H, R = 500 \YX ): + − v in ( t ) L R + − v out ( t ) (a) Circuit in “time domain” − + e V in ( j ω ) Z L ( j ω ) Z R ( j ω ) + − e V out ( j ω ) (b) Circuit in “phasor domain” Solution: The strategy is the same as the previous part, using the voltage divider formula, i.e. , e V out ( j ω ) = Z R Z R + Z L e V in ( j ω ) A similar manipulation to the previous part gives H ( j ω ) = e V out ( j ω ) e V in ( j ω ) = R R + j ω L (14) At low frequencies, we have lim ω → 0 | H ( j ω ) | = lim ω → 0 R √ R 2 + ω 2 L 2 = 1 (15) while at high frequencies, we have lim ω → ª§¦ª | H ( j ω ) | = lim ω → ª§¦ª R R 2 + ω 2 L 2 = 0 (16) So this circuit is a low-pass filter. Notice that this circuit resembles the one in the previous part, except we have replaced the capacitor with an inductor. For this transfer function, | H ( j ω ) | max = 1. Thus, to find the cutoff frequency ω c , we need to find when | H ( j ω c ) | = 1 √ 2 . | H ( j ω c ) | = 1 √ 2 (17) R p R 2 + ω 2 c L 2 = 1 √ 2 (18) R 2 + ω 2 c L 2 = 2 R 2 (19) ω c = R L (20) = 500 5 = 10 2 rad s (21) Notice that this can be observed from the transfer function itself by writing it in the following form: R R + j ω L = 1 1 + j ω R L = 1 1 + j ω ω c (22) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 4

EECS 16B Discussion 4B 2023-09-20 18:03:56-07:00 (c) RCR circuit ( R 1 = 9 k \YX , R 2 = 1 k \YX , C = 1 µF): + − v in ( t ) R 1 C R 2 + − v out ( t ) (a) Circuit in “time domain“ − + e V in ( j ω ) Z R 1 ( j ω ) Z C ( j ω ) Z R 2 ( j ω ) + − e V out ( j ω ) (b) Circuit in “phasor domain” Solution: Even though there are three components instead of two, we can still use the voltage divider formula by treating R 2 and C as a single impedance given by Z = Z C + Z R 2 , giving us Z = R 2 + 1 j ω C . This would give us e V out ( j ω ) = Z Z R 1 + Z e V in ( j ω ) (23) Then, the transfer function is H ( j ω ) = e V out ( j ω ) e V in ( j ω ) = R 2 + 1 j ω C R 1 + R 2 + 1 j ω C = 1 + j ω R 2 C 1 + j ω C ( R 1 + R 2 ) (24) At low frequencies, we have lim ω → 0 | H ( j ω ) | = lim ω → 0 q 1 + ( ω R 2 C ) 2 q 1 + ( ω C ( R 1 + R 2 )) 2 = 1 (25) while at high frequencies, we have lim ω → ª§¦ª | H ( j ω ) | = lim ω → ª§¦ª q 1 + ( ω R 2 C ) 2 q 1 + ( ω C ( R 1 + R 2 )) 2 (26) = lim ω → ª§¦ª q 1 ω 2 + ( R 2 C ) 2 q 1 ω 2 + ( C ( R 1 + R 2 )) 2 (27) = CR 2 C ( R 1 + R 2 ) = R 2 R 1 + R 2 (28) So at high frequencies, this circuit behaves like a regular voltage divider with just R 1 and R 2 , as if the capacitor had vanished. This circuit is like a combination of a low-pass filter and a voltage divider: low frequency inputs are preserved, and high-frequency signals are diminished. © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 5

EECS 16B Discussion 4B 2023-09-20 18:03:56-07:00 (d) Assuming v in ( t ) = 12 sin ( ω in t ) compute the v out ( t ) using the transfer function computed in part 1.a . Remember that R = 1 k \YX and C = 1 µF for this circuit, and assume ω in = 1000 rad s . In words, what is the effect of the transfer function in part 1. a on the magnitude and phase of the input signal? Solution: To get v out ( t ) , we must first convert v in ( t ) into phasor domain to get e V in ( j ω ) , then apply the transfer function to get e V out ( j ω ) , and then convert back to time domain to get v out ( t ) . To convert from time domain to phasor domain, we use the definition we derived previously: v in ( t ) = V 0 cos ( ω t + θ ) ↔ e V in ( j ω ) = V 0 e j θ (29) Firstly, note that sin ( x ) = cos ( x − π 2 ) , so we can write v in = 12 sin ( ω t ) as v in = 12 cos ( ω t − π 2 ) . Pattern matching with the phasor definition (with V 0 = 12 and ϕ = − π 2 ), e V in ( j ω ) = 12e − j π 2 (30) Now, we can find e V out ( j ω ) by multiplying the transfer function with the output phasor. Note that we have to evaluate the transfer function at ω = ω in = 1000 rad s since that is the input angular frequency: H ( j ω in ) = j ( 10 3 )( 10 3 )( 10 − 6 ) 1 + j ( 10 3 )( 10 3 )( 10 − 6 ) (31) = j 1 + j (32) We will write H ( j ω in ) in the form | H ( j ω in ) | e j ∠ H ( j ω in ) , so that multiplying with e V in ( j ω ) will be easier. First, to find | H ( j ω in ) | : | H ( j ω in ) | = j 1 + j = 1 √ 2 (33) Next, to find ∠ H ( j ω in ) : ∠ H ( j ω in ) = ∠ ( j ) − ∠ ( 1 + j ) = π 2 − π 4 = π 4 (34) Hence, H ( j ω in ) = 1 √ 2 e j π 4 , and e V out ( j ω in ) = H ( j ω in ) e V in ( j ω in ) = 6 √ 2e − j π 4 (35) The last step is changing back to the time domain. For this step, we can use the phasor definition in the reverse direction: v out ( t ) = 6 √ 2 cos 1000 t − π 4 (36) Contributors: • Anish Muthali. • Alex Devonport. • Nathan Lambert. • Kareem Ahmad. • Kumar Krishna Agrawal. • Nikhil Jain. © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 6

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