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Pulse and Digital Circuits

by Jntu Heroes
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www.jntuworld.com www.jwjobs.net LINEAR WAVESHAPING Introduction: If a circuit is designed with components like R ,L and C then it is called linear circuit. When sinusoidal signal is applied ,the shape of the signal is preserved at the output with or without change in the amplitude and shape. But a non-sinusoidal signal alters the output when it is transmitted through a linear circuit. The process whereby the form of non-sinusoidal signals such as step, pulse, square wave, ramp and exponential is altered by transmission through a linear network is called linear wave shaping. HIGHPASS RC CIRCUIT om Consider high pass RC circuit as shown in fig.1 below. .c Fig.1 Highpass RC circuit Sm ar tz w or ld The capacitor offers high reactance at low frequency and low reactance at high frequency. Hence low frequency components are not transmitted ,but high frequencies are with less attenuation. Therefore the output is large and the circuit is called a high pass circuit. Let us see now is, what will be the response if different types of inputs, such as, sinusoidal ,step, pulse, square wave, exponential and ramp are applied to a highpass circuit., like? (i) Sinusoidal input First consider the response of a highpass RC circuit. R R + 1 / jωC Vo = Vi Vo = Vi Let ω1 = Vo Vi R  1  R +   ωC  2 2 = R  1  R 1+    ωCR  2 1 =  1  1+    ωCR  1 CR 1 = ω  1+  1  ω 2 www.jntuworld.com 2
www.jntuworld.com www.jwjobs.net At ω = ω1 Vo Vi 1 = 2 = 0.707 om Hence, f 1 is the lower cut -off frequency of the highpass circuit. Sm ar tz w or ld .c fig.2 frequency response curve for sinusoidal input. www.jntuworld.com
www.jntuworld.com www.jwjobs.net (ii) Step input A Step voltage is defined as, Vi = 0 for t < 0 Vi = V and for t ≥ 0 The output voltage is of the form Vo = B1 + B2 e -t/τ fig.3 Step Voltage where τ = RC, the time constant of the circuit. B1 is the steady state value as t → ∞ , and hence Vo → B1 Let the final value be which we denote as Vf. Then V f = B1 om B2 is determined by the initial output voltage. At t = 0, Vi = Vo = B1 +B2 Therefore, B2 = Vi – B1 .c = Vi – V f or ld Hence the general solution is Vo = V f + ( Vi - V f ) e -t/τ − t1 0.9 = e τ t1 e τ −t τ , At t = t1, Vo (t1) = 90% of V = 0.9V Sm ar Vo (t) = Ve tz w Fall time tf: When a step is applied, the time taken for the output voltage to fall from 90%of its initial value to 10% of its initial value is the fall time. It indicates how fast the output reaches its steady state value. The output voltage at any instant of time, in highpass circuit, is given by = 1/0.9 = 1.11 t1 / τ = ln(1.11) t1 = τ ln (1.11) = 0.1 τ At t = t2, Vo (t) = 10% of V = 0.1V −t2 0.1 = e τ t2 e τ = 1/ 0.1 =10 t2 = τ ln (10) = 2.3 τ ∴fall time, tf = t2- t1 = 2.3τ - 0.1τ = 2.2 τ The lower half power frequency of the highpass circuit is 1 2πRC 1 τ = RC = 2πf 1 f 1= www.jntuworld.com
www.jntuworld.com www.jwjobs.net Fall time = tf = 2.2 τ = 2 .2 0.35 = 2πf 1 f1 Sm ar tz w or ld .c om Hence, the fall time is inversely proportional to f1, the lower cut-off frequency. www.jntuworld.com

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