Equivalent Circuits and Transfer Functions

Transcription

1 R eq isc Equialent Circuits and Transfer Functions Samantha R Summerson 14 September, Equialent Circuits eq ± Figure 1: Théenin equialent circuit. i sc R eq oc Figure : Mayer-Norton equialent circuit. Recall the steps to finding the equialent circuit alues: oc, i sc, and R eq. 1. Consider the open circuit (i = 0). Find the oltage, eq = oc.. Consider the short circuit, setting the terminal oltage to zero ( = 0). Sole for i = i sc. 3. Zero the sources (all oltage sources short-circuited and current sources open-circuited) and find the equialent resistance, R eq, looking back into the terminals. R eq = eq i sc 1

2 We can find the equialent circuit by performing any two of the three aboe steps. We can sole for the third unknown quantity using the i relation. Note that if we short-circuit our Théenin equialent circuit, the current traeling clockwise is i sc. Example 1. Find the Norton equialent circuit. For sake of completeness, we will go through all three 100Ω i 1 i i 15V ± 50Ω Figure 3: Example 1 - find the equialent circuit alues. steps. (a) Let i = 0. Using our oltage diider rule, oc = = 50Ω 50Ω 100Ω (15V ) = 5V (b) Set = 0, i.e. put a wire across the terminal. The oltage across the 50Ω resistor is now zero, meaning i must be zero. We are soling for i sc = i. Using KCL, By KVL, i 1 = i, = i sc. 100i 1 = 15V, i 1 = 3 0 A, i sc = 3 0 A. (c) Zero out the oltage source, meaning short-circuit it (replace it with a wire). From the right-hand side of the circuit, the two resistors are in parallel. Thus, ( 1 R eq = ( ) 1 3 = 100 = Ω ) 1 Example. Find the equialent circuit alues.

3 100 3 Ω 5V ± Figure 4: Example 1 - Théenin equialent circuit. 3 0 A Ω oc Figure 5: Example 1 - Mayer-Norton equialent circuit. i. First, set i = 0. This implies that the current through the 10Ω resistor, which is the resistor we are measuring the output oltage across, is i 4. So, oc = = 10i 4. We only need to sole for i 4. Now we can write our KCL and KVL equations. By KVL, we hae By KCL we hae: 10 = 0i i 16i 4, 10 = 0i 1 5i 3. i 3 i 5 = i 1. First we will sole for i 1 in terms of i 4. i 1 = i i 3, i 1 = i 4, i 4 = 1 i 5, i 1 = i i 3 = (i 4 1) i 3 16i 4 = 5i 3 from KVL equations 16 5 i 4 = i 3 i 1 = i i 4 = 1 5 i 4 1 3

4 0Ω i 1 i i 3 i 4 6Ω i 10V ± 5Ω 1A 10Ω i 5 Figure 6: Example - find the equialent circuit alues. Using this expression for i 1, we plug it into our first KCL equation and sole for i 4. Now, 10 = 0( 1 5 i 4 1) 16i 4 = 100i 4 0 i 4 = 3 10 A oc = 10i 4 = 3V. ii. Put a wire across the terminals, i.e. set = 0. We are soling for i sc = i. Due to the short circuit, no current will go through the 10Ω resistor. Thus, i sc = i = i 4. From KVL, we hae 10 = 0i 1 6i 4, 10 = 0i 1 5i 3. These two equations together imply 6 5 i 4 = i 3. From KCL, we hae i 1 = i i 3, 1 i = i 4, i = i 4, i = 1 i 5, i 3 i 5 = i 1. Combining the first two equations, we hae i 1 = (i 4 1) i 3, = i i 4, from the KVL equations = 11 5 i 4 1. Using this alue for i 1 in the first KVL equation, ( ) = 0 5 i 4 1 6i 4, = 50i 4 0, i 4 = 3 5 A. iii. By the i relation, R eq = oc i sc = = 5Ω. 4

5 Equialent Capacitance and Inductance.1 Capacitors in Series and Parallel Consider two capacitors in parallel. The oltage across the capacitors will be the same,, though the current will be different. By KCL, the current going into the node is the equal to the sum of the current through both capacitors: i = i 1 i. Using our i relation for capacitors, we can re-write this. i = i 1 i, d = C 1 dt C d = (C 1 C ) d = C eq d dt. Thus, the equialent capacitance for two capacitors in parallel is C eq = C 1 C. This is analogous to the rule for resistors in series. Consider two capacitors in series. In this case, the current through both is the same but the oltages are different. Howeer, the oltage across both has the relation = 1. Since differentiating is a linear operation, we can take the deriatie of both sides and plug in the i relations. d dt = d 1 dt d = i i, C 1 C ( 1 = 1 ) i, C 1 C i =. C eq Thus, the equialent capacitance for two capacitors in series is ( 1 C eq = 1 ) 1. C 1 C This is analogous to the rule for resistors in parallel.. Inductors in Series and Parallel We can do similar analysis for inductors. If we consider two inductors in parallel, the oltage across both will be the same,, though the current will be different. Thus current going into the node connecting the two inductors has the following relation: i = i 1 i. 5

6 We can take the deriatie of both sides and use our i relations. di dt = di 1 dt di =, L 1 L ( 1 = 1 ), L 1 L =. L eq The equialent inductance for two inductors in parallel is ( 1 L eq = 1 ) 1. L 1 L This is analogous to the rule for resistors in parallel. Finally, consider two inductors in series. The current through both inductors will be the same, i. The oltage across the two is equal to the sum of the oltages, Using our i relations, we hae = 1. = 1, di = L 1 dt L di = (L 1 L ) di = L eq di dt. The equialent inductance for two inductors in series is L eq = L 1 L. This is analogous to the rule for resistors in series. The rules for inductors are similar to those for resistors, whereas the rules for capacitors are the opposite. 3 Transfer Function For an LTI system, for any sinusoidal input, the output will be a sinusoid with the same frequency but possibly different amplitude and phase. Consider an input oltage in (t) = V in e jπft. The output oltage for the system will hae the form out (t) = V out e j(πftφ). 6

7 We can write this output oltage in terms of the input oltage. out (t) = V out V in V in e jπft e jφ, = V out V in e jφ V in e jπft, = V out V in e jφ in (t). The term scaling the input oltage is called the transfer function, H. Definition 1. The transfer function for an LTI system is defined as H(f) = out in. The transfer function defines the response of the system to any complex exponential input. It essentially defines the system. Example 3. Consider a circuit/system where out (t) = in (t M). M is a constant. Find H(f). First of all, we need to check if the system is LTI. Since it is a simple delay system, it is. Now we consider a complex exponential input, The corresponding output is in = e jπft. out = e jπf(tm). Using the formula for the transfer function, we hae H(f) = ejπf(tm) e jπft, = e jπfm. Since the magnitude of this filter is always one, it is an all-pass filter. Example 4. Consider an LTI system with transfer function H(f) and in (t) = cos(πft). What is out in terms of H(f)? Like we do so many time, we re-write the cosine as a sum of exponentials (Euler s formula) Since in (t) = 1 ( e jπft e jπft) e jπft H(f)e jπft, e jπft H(f)e jπft, and our system is LTI (i.e. superposition holds), we hae out (t) = 1 H(f)ejπft 1 H(f)ejπft. 7

8 From class we learned the two main properties of the magnitude and phase of the transfer function. Namely, The magnitude of H is an een function, i.e. The phase of H is an odd function, i.e. From class, we also saw that for a real input the output always has the form H(f) = H(f). H(f) = H(f). in (t) = cos(πft), out (t) = H(f) cos(πft H(f)). Example 5. Consider an LTI system with input and transfer function in (t) = 1 cos(πt) cos(πt), H(f) = { e jπft πf < 4 0 otherwise. Find out. First we note that the input is a sum of three cosine functions at frequences f = 0, 1, 1. By supersition and our deriation in class, out = H(0) cos(0) H( 1 ( )) 1 (πt ) cos H H(1) cos(πt H(1)). By examination, we see that Thus, H(0) = 1, ( ) 1 H = e jπ, H(1) = 0. out = 1 e jπ cos(πt e jπ ) 0, = 1 cos(πt π). We can see that the highest frequency component of the input, cos(πt), was filtered out (i.e. that frequency does not appear in the output). This makes sense, as our filter is a low-pass filter. 8