dis04B_sol

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16B

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Electrical Engineering

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Oct 30, 2023

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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

dis04B_sol

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