LCS_4_notes - 1 LINEAR CIRCUITS AND SIGNALS Frequency...

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1 LINEAR CIRCUITS AND SIGNALS Frequency Response Department of Electrical and Computer Engineering University of Miami I. S OME E XAMPLES Suppose a dissipative circuit is being driven by an independent sinusoidal voltage or current source, e.g., v i ( t ) = Re [ V i e jωt ] = | V i | cos[ ωt + V i ]; i i ( t ) = Re [ I i e jωt ] = | I i | cos[ ωt + I i ] . (1) Here V i and I i denote the phasors corresponding to v i ( t ) and i i ( t ) , respectively. Recall that we may use these phasors to determine the system response at steady-state. _ + v i (t) R + _ v o (t) 1/j ω C (a) RC circuit. -5 -4 -3 -2 -1 0 1 2 3 4 -2 -1 0 1 2 ω rad/s H( ω ) rad -5 -4 -3 -2 -1 0 1 2 3 4 0 0.2 0.4 0.6 0.8 1 ω rad/s |H( )| (b) Frequency response. Fig. 1. Frequency response of the RC circuit. Example For the RC circuit in Figure 1(a), we have V o = R R + 1 /jωC V i = jωRC 1 + jωRC V i .
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2 Therefore v o ( t ) = | H ( ω ) || V i | cos[ ωt + V i + H ( ω )] , where H ( ω ) = V o V i = jωRC 1 + jωRC = ⇒ | H ( ω ) | = | ω | RC p 1 + ω 2 ( RC ) 2 ; H ( ω ) = + π/ 2 - tan - 1 [ ωRC ] , for ω 0; - π/ 2 - tan - 1 [ ωRC ] , for ω 0 . Note that H ( ω ) = monotonically decreasing from + π/ 2 to 0 , for ω 0; monotonically increasing from - π/ 2 to 0 , for ω 0 . Let us substitute some frequency values: ω = 0 : H (0) = 0 = ⇒ | H (0) | = 0 = v o ( t ) = 0 , i.e., the output is completely suppressed at DC. ω → ±∞ : H ( ±∞ ) = 1 = ⇒ | H ( ±∞ ) | = 1; H ( ±∞ ) = 0 = v o ( t ) = | V i | cos[ ωt + V i ] = v i ( t ) , i.e., the input propagates to the output with no change at high frequencies. ω = +1 /RC , i.e., the positive ‘break’ frequency (which corresponds to the circuit time constant): H (+1 /RC ) = j 1 + j = ⇒ | H (+1 /RC ) | = 1 2 ; H (+1 /RC ) = + π 4 = v o ( t ) = 1 2 | V i | cos[ ωt + V i + π/ 4] . ω
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LCS_4_notes - 1 LINEAR CIRCUITS AND SIGNALS Frequency...

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