ECEN 60 Circuits/Electronics Spring 007-7-07 P. Mathys Review of st Order Circuit Analysis First Order Differential Equation Consider the following circuit with input voltage v S (t) and output voltage v O (t). i R (t) R i C (t) v S (t) C v O (t) Recall that for capacitors i C (t) = C v () C (t), where v() C (t) denotes the first derivative (with respect to t) of v C (t). Using KCL i R (t) = i C (t) = Thus, the differential equation for this circuit is v S (t) v O (t) R = C v () O (t). v () O (t) v O(t) = v S (t), with initial condition v O (0) = V 0. Another way to write this is v () O (t) = v S(t) v O(t). This can be visualized using a block diagram with an integrator as shown below. v S (t) v () O (t) V 0 v O(t)
Natural Response To obtain the natural response (or homogeneous solution) of the st order differential equation v () O (t) v O(t) = v S (t), set v S (t) = 0 and assume a solution of the form v N (t) = K e st, t 0, = v () N (t) = Ks est, t 0, for some constants K and s. Substitute in the differential equation to obtain The nontrivial solution is Ks e st K e st = K (s ) e st = 0. s = 0 = s = = v N (t) = K e t/, t 0. Because the voltage across a capacitor is a state variable, the initial condition V 0 is the voltage across the capacitor at time t = 0, i.e., V 0 = v O (0) = v N (0) = K = v N (t) = V 0 e t/, t 0. 3 Forced Response: Step Input Let v S (t) = u(t) and assume a forced response (or particular solution) of the form v F (t) = V F, t 0, = v () F (t) = 0, t 0. Substitute in the differential equation to obtain V F = = v F (t) =, t 0. Next use superposition to combine the natural and forced solutions as v O (t) = v N (t) v F (t) = K e t/, t 0. Then use the initial condition V 0 = v O (0) to solve for K This yields the complete solution as V 0 = v O (0) = K = K = V 0. v O (t) = (V 0 ) e t/ }{{} V }{{} A natural forced = V 0 e }{{ t/ } ZIR ( e t/, t 0. }{{} ZSR The natural solution is also called transient response and the forced solution is also called steady-state response. ZIR stands for zero-input response and ZSR stands for zero-state response. The following graph shows v O (t) when V 0 /3.
V 0 0 0 3 t 4 Forced Response: Sinusoidal Input Let v S (t) = cos ωt, t 0 and assume a solution of the form v F (t) = V F cos(ωt φ) = a cos ωt b sin ωt, where the Fourier coefficients a, b are given by (from cos(α β) = cos α cos β sin α sin β) a = V F cos φ, b = V F sin φ, and thus V F = a b, φ = tan b a. The first derivative of v F (t) (in Fourier coefficient format) is v () F (t) = aω sin ωt bω cos ωt, t 0. After substituting in the differential equation one obtains (aω sin ωt bω cos ωt) a cos ωt b sin ωt = cos ωt, t 0. Since cos ωt and sin ωt are orthogonal functions, this yields the following two equations (one for the coefficents of all the cos ωt terms and one for all the coefficients of the sin ωt terms) a b ω =, and a ω b = 0. In matrix-vector form this is [ ] [ ] ω a = ω b [ ] VA 0. Using Kramer s rule to solve for a and b yields ω 0 a = ω ω = (ω), 3
and b = ω 0 ω ω = ω (ω). Using superposition, the solution for v O (t) is the sum of v N (t) and v F (t), i.e., v O (t) = K e t/ To solve for K the initial condition v O (0) = V 0 is used (cos ωt ω sin ωt), t 0. (ω) V 0 = K = K = V (ω) 0 (ω). Thus, the complete solution is ( V ) A v O (t) = V 0 e t/ (cos ωt ω sin ωt) (ω) }{{ (ω) }}{{} transient response steady-state response. 5 Frequency Response More physical insight can be gained by converting the steady-state response from the Fourier coefficient format back to the form V F cos(ωt φ). This results in V F = a b = (ω) and φ = tan b a = tan ω, and thus ( V ) A v O (t) = V 0 e t/ (ω) }{{ } transient response (ω) cos(ωt tan ω) } {{ } steady-state response The combination of V F and φ is called the frequency response of the circuit. Individually, V F is called the magnitude of the frequency response and φ is called the phase of the frequency response. If a circuit has a frequency response magnitude that is larger for lower frequencies than higher frequencies, it is called a lowpass filter (LPF). Conversely, if a circuit has a frequency response magnitude that is smaller for lower frequencies than higher frequencies, it is called a highpass filter (HPF). Of special importance is the (cutoff) frequency ω c at which the output power of the circuit is reduced to one half of the input power. For V F as given above, this occurs for ω c = / as can be easily seen from V F ω=/ = (/) = = 0.707.. 4
Note that at ω = ω c φ ω=/ = tan (/) = tan = 45. 6 Phasors Using Euler s relationship e jθ = cos θ j sin θ, a sinusoid of the form v(t) = cos(ωt φ) can be expressed as The quantity v(t) = cos(ωt φ) = Re{ e j(ωtφ) } = Re{ e }{{ jφ e } jωt }. =V V = e jφ is called the (complex-valued) phasor representation of v(t). This can be expressed symbolically as v(t) V. Linearity (and thus the superposition principle) holds for phasors, i.e., K v (t) K v (t) K V K V, for arbitrary real constants K and K. Note that v (t) and v (t) must both be sinusoids of the same frequency for this to hold. The phasor representation of the derivative of v(t) is obtained from and thus v () (t) = d dt Re{V ejωt } = Re { d dt V ejωt} = Re{jωV e jωt }, v () (t) jωv. The i(t)-v(t) characteristic of a resistor can be expressed as For capacitors one obtains v R (t) = R i R (t) V R = R I R = Z = R. i C (t) = C v () C (t) I C = V C = Z =. For inductors the relationships are v L (t) = L i () L (t) V L = jωl I L = Z = jωl. 5
7 Phasor Analysis The figure below shows the LPF circuit labeled for phasor analysis. R V O Using voltage division one obtains immediately V O = R = jω. Multiplying through by jω yields the relationship jω V O V O = v () O (t) v O(t) = v S (t), i.e., the differential equation can be obtained directly from the phasor analysis (and vice versa). Dividing the output phasor V O = V O e jφ O by the input phasor = e jφ S, one can define the transfer function H of the circuit as H = V O = V O e jφo V s e jφ S = V O e j(φ Oφ S ). Note that H is a complex quantity in general that can be written as H = H e j H. The quantity H = V O / is the magnitude of the frequency response and H = φ O φ S is the phase of the frequency response of the circuit. For the LPF circuit H = jω = jω (ω) = (ω) ej tan ω, where H = ( (ω) ) is the magnitude and H = tan ω is the phase of the frequency response of the circuit. This is the same result as found earlier from the forced solution of the differential equation for a sinusoidal input signal. 6
8 Analysis of Highpass Filter If the roles of the capacitor and the resistor are interchanged, then the circuit shown below is obtained. C i C (t) v C (t) i R (t) v S (t) R v O (t) Because of the capacitor in series with the input signal v S (t), this circuit does not pass dc, but passes high frequencies and is therefore a highpass filter or differentiator. Starting from the phasor analysis this time, the schematic is relabeled as follows: V C R V O Using voltage division yields immediately V O = R R = which translates into the differential equation jω jω, jω V O V O = jω v () O (t) v O(t) = v () S (t). The frequency response of the circuit is found as H = V O = jω jω = jω ( jω) (ω) = ω (j ω). (ω) The magnitude H and the phase H of the frequency response are therefore H = ω, H = (ω) tan ω. The half-power (or cutoff) frequency is ω c = / and H ωc = / and H ωc = tan = 45. 7
To derive the step response of the system with a specified initial condition, the form of the differential equation obtained above is not suitable because v O (t) is not a state variable of the system. In this case it is better to first derive a differential equation for the state variable v C (t) = v S (t) v O (t), and then to obtain v O (t) as a function of v C (t). Noting that v O (t) = v S (t) v C (t) and i C (t) = i R (t) one can write v O (t) = v S (t) v C (t) = R i C (t) = v () C (t) = v() C (t) v C(t) = v S (t), with initial condition v C (0) = V 0. Since i C (t) = i R (t), the output voltage is related to v C (t) by v O (t) = R i C (t) = v () C (t). This is shown in block diagram form in the following figure: v O (t) v S (t) v () C (t) V 0 v C (t) The natural solution of the differential equation for v C (t) is v N = K e t/, t 0, and the forced solution in response to a step input v S (t) = u(t) is v F (t) =, t 0. Combining v N (t) and v F (t) and using the initial condition v C (0) = V 0 to solve for K yields v C (t) = (V 0 ) e t/, t 0. Thus, v O (t) = v () C (t) in response to a step input of amplitude is v O (t) = ( V 0 ) e t/ = e }{{ t/ V } 0 e t/, t 0. }{{} ZSR ZIR c 003 007, P. Mathys. Last revised: -3-07, PM. 8