# Modeling the Transient Response of Thermal Circuits
## 4 Analytical Modeling of a Complete Thermal System
### Transient Response Model  
To determine the transient response model, each thermal resistance of the steady-state model was converted into a thermal resistance and heat capacitor in parallel. The new circuit is depicted in Figure 6c. Using the [PERSON]'s circuit laws, one can write the following eight simultaneous equations:  
\[Q =q_{1}+q_{2},\] (64) \[q_{1} =q_{5}+q_{6},\] (65) \[q_{2} =q_{7}+q_{8},\] (66) \[q_{2} =q_{3}+q_{4},\] (67) \[\frac{dq_{5}}{dt} =\frac{1}{R_{1}C}q_{6}  
Then, by differentiating the equations, we have:  
\[u_{3}+\frac{dq_{4}}{dt} =-u_{5}-\frac{dq_{6}}{dt}, \tag{75}\] \[u_{7}+\frac{dq_{8}}{dt} =u_{3}-\frac{dq_{4}}{dt},\] (76) \[\frac{du_{5}}{dt} =\frac{1}{R_{1}\mathrm{C}}\frac{dq_{6}}{dt},\] (77) \[\frac{du_{3}}{dt} =\frac{1}{R_{2}\mathrm{C}}\frac{dq_{4}}{dt},\] (78) \[\frac{du_{7}}{dt} =\frac{1}{R_{3}\mathrm{C}}\frac{dq_{8}}{dt},\] (79) \[\frac{dq_{6}}{dt} =\frac{dq_{4}}{dt}+\frac{dq_{8}}{dt}. \tag{80}\]  
Using Equations (75), (76) and (80), and grouping them into a single system, one arrives at the following system of three differential equations in the matrix notation:  
\[ \begin{bmatrix}\frac{du_{4}}{dt}\\ \frac{du_{5}}{dt}\\ \frac{du_{7}}{dt}\end{bmatrix} =M \begin{bmatrix}u_{4}\\ u_{5}\\ u_{7}\end{bmatrix} , \tag{81}\]  
where  
\[M= \begin{bmatrix}-\frac{2}{3\mathrm{C}R_{2}}&-\frac{1}{3\mathrm{C}R_{2}}&\frac {1}{3\mathrm{C}R_{2}}\\ -\frac{1}{3\mathrm{C}R_{1}}&-\frac{2}{3\mathrm{C}R_{1}}&-\frac{2}{3\mathrm{C}R _{1}}\\ \frac{1}{3\mathrm{C}R_{3}}&-\frac{1}{3\mathrm{C}R_{3}}&-\frac{2}{3\mathrm{C}R _{3}}\end{bmatrix} . \tag{82}\]  
These three differential equations are coupled, but they are linear and of first order. Therefore, it suffices to obtain the eigenvalues of the matrix \(M\) to calculate the time constants. In this case, there are three time constants:  
\[det \begin{bmatrix}-\frac{2}{3\mathrm{C}R_{2}}-\lambda&-\frac{1}{3\mathrm{C}R_{ 2}}&\frac{1}{3\mathrm{C}R_{2}}\\ -\frac{1}{3\mathrm{C}R_{1}}&-\frac{2}{3\mathrm{C}R_{1}}-\lambda&-\frac{2}{3 \mathrm{C}R_{1}}\\ \frac{1}{3\mathrm{C}R_{3}}&-\frac{1}{3\mathrm{C}R_{3}}&-\frac{2}{3\mathrm{C}R _{3}}-\lambda\end{bmatrix} =0. \tag{83}\]  
By doing the appropriate calculations, one arrives at the time constants:  
\[\lambda_{0} =0, \tag{84}\] \[\lambda_{1} =-\lambda^{\prime}-\lambda^{\prime\prime},\] (85) \[\lambda_{2} =-\lambda^{\prime}+\lambda^{\prime\prime}, \tag{86}\]  
where  
\[\lambda^{\prime} =-\frac{1}{3\mathrm{C}R^{*}}, \tag{87}\] \[\lambda^{\prime\prime} =-\frac{1}{3\mathrm{C}}\sqrt{\frac{1}{R^{**}}-\frac{1}{R^{***}}}, \tag{88}\]\[\frac{1}{R^{*}} =\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}, \tag{89}\] \[\frac{1}{R^{**}} =\frac{1}{R_{1}^{2}}+\frac{1}{R_{2}^{2}}+\frac{1}{R_{3}^{2}},\] (90) \[\frac{1}{R^{***}} =\frac{1}{R_{1}R_{2}}+\frac{1}{R_{2}R_{3}}+\frac{1}{R_{1}R_{3}}. \tag{91}\]  
Let us suppose that \(T\), \(V\), \(W\) are the eigenvectors associated with the eigenvalues \(\lambda_{0}\), \(\lambda_{1}\) and \(\lambda_{2}\), respectively. Then, the solution is  
\[ \begin{bmatrix}u_{3}\\ u_{5}\\ u_{7}\end{bmatrix} =AT+BVe^{\lambda_{1}t}+DWe^{\lambda_{3}t}. \tag{92}\]  
The sum of two exponential functions is never equal to one exponential. However, for certain conditions, it is possible to approximate the sum to one exponential function. In fact, \(Ae^{\lambda_{1}t}+Be^{\lambda_{2}t}=Ae^{-\lambda^{2}t}e^{-\lambda^{2}t}+Be^{- \lambda^{2}t}e^{\lambda^{2}t}=e^{-\lambda^{2}t}(Ae^{-\lambda^{2}t}+Be^{\lambda ^{2}t})\). If \(t\gtrsim\lambda^{\prime\prime}\), then \(Ae^{\lambda_{1}t}+Be^{\lambda_{2}t}\simeq e^{-\lambda^{2}t}(Be^{\lambda^{2}t} )=Be^{-(\lambda^{2}-\lambda^{2})t}=Be^{-\lambda_{2}t}\). To calculate \(T_{cold}\) and \(T_{hot}\), one needs \(q_{5}\) and \(q_{7}\).