Question 6.2: Computing the Frequency Response of a Circuit Compute the fr...

Known Quantities: R_{1}=1  \mathrm{k} \Omega ; L=2  \mathrm{mH} ; R_{L}=4  \mathrm{k} \Omega.

Find: The frequency response H_{Z}(j \omega)=\mathbf{V}_{L}(j \omega) / \mathbf{I}_{S}(j \omega).

Assumptions: None.

Analysis: To determine expressions for the load voltage, we recognize that the load current can be obtained simply by using a current divider between the two branches connected to the current source, and that the load voltage is simply the product of the load current times R_{L}.

Using the current divider rule, we obtain the following expression for \mathbf{I}_{L} :

\mathbf{I}_{L}=\frac{\frac{1}{R_{L}+j \omega L}}{\frac{1}{R_{L}+j \omega L}+\frac{1}{R_{1}}} \mathbf{I}_{S}=\frac{1}{1+\frac{R_{L}}{R_{1}}+j \frac{\omega L}{R_{1}}} \mathbf{I}_{S}

and

\frac{\mathbf{V}_{L}}{\mathbf{I}_{S}}(j \omega)=H_{Z}(j \omega)=\frac{I_{L} R_{L}}{I_{S}}=\frac{R_{L}}{1+\frac{R_{L}}{R_{1}}+j \frac{\omega L}{R_{1}}}

Substituting numerical values, we obtain:

H_{Z}(j \omega)=\frac{4 \times 10^{3}}{1+4+j \frac{2 \times 10^{-3} \omega}{10^{3}}}=\frac{0.8 \times 10^{3}}{1+j 0.4 \times 10^{-6} \omega}

Comments: You should verify that the untis of the expression for H_{Z}(j \omega) are indeed ohms, as they should be from the definition of H_{Z}.

Focus on Computer-Aided Tools: A computer-generated solution of this problem may be found in the CD-ROM that accompanies this book.