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Berkeley ELENG 105 - Discussion Notes

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EE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 1 Monday, March 28, 2005 EE 105 Discussion Section 101 Announcements Response of Linear Circuits to Sinusoidal Input Signals A linear circuit is a circuit that consists of resistors, capacitors, inductors, voltage sources (independent and/or dependent), and/or current sources (independent and/or dependent). Now, suppose that a linear circuit is excited by a sinusoidal input signal (either current or voltage). In other words, suppose that the input to a linear circuit is )cos()(φω+⋅⋅= tAtx (1) where, )(sin)(sinsinradinusoidalinputofoffsetphasesradinusoidalinputoffrequencyusoidalinputofamplitudeA===φω Voltage and current signals will be generated in the linear circuit in response to the aforementioned sinusoidal input. These voltage and current signals are also sinusoidal in nature. In fact, the frequencies of the generated voltage and current signals is identical to the frequency of the input signal. Hence, one can mathematically characterize each voltage and current signal generated by the input sinusoid as follows )cos()(θω+⋅⋅= tBty (2) where, )(/)(//radincurrentvoltagegeneratedofoffsetphasesradincurrentvoltagegeneratedoffrequencycurrentvoltagegeneratedofamplitudeB===θω There are various methods that one can employ in order to calculate B and θ. Here, we will focus on using the Laplace Transform. Using the Laplace Transform to Find the Sinusoidal Response of a Linear Circuit For the theory behind using the Laplace Transform as it pertains to circuit/system analysis, please see EE 20/120. Here, we will simply use the results by following certain steps. To also demonstrate the process, we will carry these steps out using the circuit in Figure 1 as an example.EE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 2 1. Represent every current and voltage in the circuit using Laplace notation. Carrying this out on the circuit in Figure 1 results in the circuit depicted in Figure 2. 2. In the Laplace domain, the concept of impedance (i.e.Z) is used rather than resistance (i.e.R) to describe Ohm’s Law. More specifically, the following transformation takes place in the Laplace domain, )()()()()( sIsZsVbecomestiRtv⋅=⋅= (3) Like resistance, impedance has units of ohms. The impedance formulas for the various possible circuit elements are as follows, Impedance of a resistor with resistance RsZRR==)( (4) Impedance of a capacitor with capacitance CssZCC⋅==1)( (5) Impedance of an inductor with inductance LssZLL⋅==)( (6) )(3ti )(2ti)(1ti )(tvX1L )(tiS 1C1RFigure 1. Parallel RLC circuit )(3sI )(2sI)(1sI)(sVX1L )(sIS 1C1RFigure 2. Representing signals using Laplace notationEE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 3 Hence, in step 2, replace each resistor, capacitor, and inductor in the circuit with its corresponding impedance. Carrying this out on the circuit in Figure 2 results in the circuit depicted in Figure 3. 3. Use the circuit calculation techniques learned in EE 40 (node voltage, mesh current, voltage divider, current divider, Thèvenin and Norton equivalents, etc.) to obtain s-domain expressions for all of the voltages and currents of interest. Upon doing so, find the s-domain transfer function that relates each voltage and current of interest to the applied input. More specifically, )()()( sXsHsYinputfunctiontransferoutput⋅=⇒⋅= (7) Let us, then, try to find appropriate expressions for )(),(),(),(321sIsIsIsVX and their respective transfer functions. The equivalent impedance to the right of the input )(sIS is as follows, 11111211111)(//1//)(RLsRLCsRLssZLsCsRsZeqeq+⋅+⋅⋅⋅⋅⋅=∴⋅⋅= (8) Thus, 11111211111111211)()()()()()()(RLsRLCsRLssHsIsVsIRLsRLCsRLssIsZsVSXSSeqX+⋅+⋅⋅⋅⋅⋅==∴⋅+⋅+⋅⋅⋅⋅⋅=⋅= (9) 11111212111111121111)()()()(1)()(RLsRLCsLssHsIsIsIRRLsRLCsRLsZsVsISSRX+⋅+⋅⋅⋅⋅==∴⋅⋅+⋅+⋅⋅⋅⋅⋅== (10) )(3sI )(2sI)(1sI)(sVX1Ls ⋅)(sIS 11Cs ⋅1RFigure 3. Complete Laplace representation of original circuitEE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 4 11111211123211111121112)()()()()()(RLsRLCsRLCssHsIsIsICsRLsRLCsRLsZsVsISSCX+⋅+⋅⋅⋅⋅⋅⋅==∴⋅⋅⋅+⋅+⋅⋅⋅⋅⋅== (11) 11111214311111121113)()()()(1)()(RLsRLCsRsHsIsIsILsRLsRLCsRLsZsVsISSLX+⋅+⋅⋅⋅==∴⋅⋅⋅+⋅+⋅⋅⋅⋅⋅== (12) 4. Using the transfer function, one can compute the B and θ (see (2)) for each voltage/current of interest as follows, AjsHB ⋅⋅==signalinput theof frequency at the evaluated)(ω (13) φωθ+⋅=∠=signalinput theof frequency at the evaluated)( jsH (14) Example Let the frequency of the input signal )(tiSbe sradekHz 328.612 =⋅⋅=πω, its amplitude 1=A , and its phase 0=φ. Let’s also assume that mHLpFCkR 25,1,1111==Ω= . Then, using (13) and (14) yields ()°+⋅⋅=∴°=⋅=∠=⋅=81)328.6(sin1551.0)(81)328.6(1551.0)328.6(122tsradetiejsHejsH ()°−⋅⋅=∴°−=⋅=∠=⋅=9.8)328.6(sin9879.0)(9.8)328.6(9879.0)328.6(344tsradetiejsHejsH More on Transfer Functions Every transfer function can be represented as a ratio of two polynomials. More specifically, )()()(sQsPsH= (15) where,EE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 5 )(),( sQsP are each polynomial functions of s . The roots of )(sP and )(sQ have special names. More specifically, the roots of )(sP are referred to as the zeros of the transfer function )(sH . The roots of )(sQ , on the other hand, are referred to as the poles of the transfer function )(sH . Common Transfer Functions The circuits depicted in Figure 4 occur frequently in circuit analysis/design. Hence, it is useful to have their associated transfer functions memorized. It can be shown that 111111)()(CRsRsIsV⋅⋅+= (16) 222311)()(CRssVsV⋅⋅+= (17) 2C 2R)(2sV)(1sV )(1sI 1C1R )(3sV Figure 4. Common


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Berkeley ELENG 105 - Discussion Notes

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