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ssrs barcodelib V0 est e Rt=L u t R Ls in Software
V0 est e Rt=L u t R Ls Decoding QR In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Code ISO/IEC18004 Generation In None Using Barcode creation for Software Control to generate, create Quick Response Code image in Software applications. 17d QR Code ISO/IEC18004 Reader In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. QR Code ISO/IEC18004 Drawer In C# Using Barcode drawer for .NET Control to generate, create Quick Response Code image in Visual Studio .NET applications. Special Case. If the forcing function has the same exponent as that of the natural response s R=L , the forced response needs to be ip t I0 te Rt=L . This can be veri ed by substitution in (16), which also yields I0 V0 =L The natural response is the same as (17b). The total response is then i t ip t ih t I0 t A e Rt=L From i 0 i 0 0 we nd A 0, and so i t I0 te Lt=R u t , where I0 V0 =L. QRCode Drawer In VS .NET Using Barcode creator for ASP.NET Control to generate, create Quick Response Code image in ASP.NET applications. Print QR Code ISO/IEC18004 In .NET Framework Using Barcode generation for .NET framework Control to generate, create Denso QR Bar Code image in VS .NET applications. CHAP. 7] QR Code ISO/IEC18004 Generation In VB.NET Using Barcode creator for Visual Studio .NET Control to generate, create QR Code image in .NET framework applications. ECC200 Creator In None Using Barcode generator for Software Control to generate, create Data Matrix ECC200 image in Software applications. FIRSTORDER CIRCUITS
USS Code 39 Maker In None Using Barcode creation for Software Control to generate, create USS Code 39 image in Software applications. EAN13 Creation In None Using Barcode maker for Software Control to generate, create UPC  13 image in Software applications. RESPONSE OF RC AND RL CIRCUITS TO SUDDEN SINUSOIDAL EXCITATIONS
UPCA Encoder In None Using Barcode generator for Software Control to generate, create UPC A image in Software applications. Drawing Barcode In None Using Barcode generation for Software Control to generate, create barcode image in Software applications. When a series RL circuit is connected to a sudden ac voltage vs V0 cos !t (Fig. 719), the equation of interest is Ri L The solution is i t ih ip where ih t Ae Rt=L and ip t I0 cos !t di V0 cos !t u t dt 18 Generating RM4SCC In None Using Barcode drawer for Software Control to generate, create RoyalMail4SCC image in Software applications. EAN13 Supplement 5 Recognizer In Java Using Barcode reader for Java Control to read, scan read, scan image in Java applications. By inserting ip in (18), we nd I0 : V0 I0 p R2 L2 !2 Then and tan 1 L! R Draw Bar Code In C#.NET Using Barcode drawer for .NET Control to generate, create bar code image in .NET applications. DataMatrix Generation In .NET Framework Using Barcode encoder for ASP.NET Control to generate, create ECC200 image in ASP.NET applications. i t Ae Rt=L I0 cos !t
GS1  12 Creation In ObjectiveC Using Barcode printer for iPhone Control to generate, create UPC A image in iPhone applications. Print Barcode In Java Using Barcode creation for Android Control to generate, create bar code image in Android applications. From i 0 0, we get A I0 cos . Therefore, i t I0 cos !t cos e Rt=L
Code 128A Recognizer In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. GTIN  128 Creation In Java Using Barcode generator for Android Control to generate, create UCC128 image in Android applications. Fig. 719 SUMMARY OF FORCED RESPONSE IN FIRSTORDER CIRCUITS Consider the following di erential equation: dv t av t f t dt 19 The forced response vp t depends on the forcing function f t . Several examples were given in the previous sections. Table 72 summarizes some useful pairs of the forcing function and what should be guessed for vp t . The responses are obtained by substitution in the di erential equation. By weighted linear combination of the entries in Table 72 and their time delay, the forced response to new functions may be deduced. FIRSTORDER ACTIVE CIRCUITS
Active circuits containing op amps are less susceptible to loading e ects when interconnected with other circuits. In addition, they o er a wider range of capabilities with more ease of realization than passive circuits. In our present analysis of linear active circuits we assume ideal op amps; that is; (1) the current drawn by the op amp input terminals is zero and (2) the voltage di erence between the inverting and noninverting terminals of the op amp is negligible (see 5). The usual methods of analysis are then applied to the circuit as illustrated in the following examples. FIRSTORDER CIRCUITS
[CHAP. 7
Table 72 f t 1 1 a t 1 a a2 est s a te at A cos !t where 1 A p 2 !2 a and tan ! a tan ! a b vp t t est ; s 6 a e at cos !t
e bt cos !t
Ae bt cos !t
where
1 A q a b 2 !2 EXAMPLE 7.15 Highpass lter. The op amp in the circuit of Fig. 744 is ideal. Find the unitstep response of the circuit; that is, v2 for v1 u t : The inverting input terminal of the op amp is at virtual ground and the capacitor has zero voltage at t 0 . The 1V step input therefore generates an exponentially decaying current i through R1 C (from left to right, with a time constant R1 C and initial value of 1=R1 ). i 1 t= R1 C e u t R1 All of the preceding current passes through R2 (the op amp draws no current), generating v2 R2 i at the output terminal. The unitstep response is therefore v2 R2 t= R1 C e u t R1 EXAMPLE 7.16 In the circuit of Fig. 744 derive the di erential equation relating v2 to v1 . Find its unitstep response and compare with the answer in Example 7.15. Since the inverting input terminal of the op amp is at virtual ground and doesn t draw any current, the current i passing through C, R1 , and R2 from left to right is v2 =R2 . Let vA be the voltage of the node connecting R1 and C. Then, the capacitor voltage is v1 vA (positive on the left side). The capacitor current and voltage are related by v2 d v1 vA R2 dt To eliminate vA , we note that the segment made of R1 , R2 , and the op amp form an inverting ampli er with v2 R2 =R1 vA , from which vA R1 =R2 v2 . Substituting for vA , we get v2 R 1 C dv2 dv R2 C 1 dt dt To nd the unitstep response, we rst solve the following equation: dv R2 C t>0 v2 R 1 C 2 0 t<0 dt The solution of the preceding equation is R2 C 1 e t= R1 C u t . derivative of the preceding solution. The unitstep response of the circuit is the time CHAP. 7]

