### Phase Shift Due to Finite Length of Circuit ConductorsOn a particular type of circuit board, assume a signal travels at \( \frac{2}{3} \) of the speed of light. Suppose a high-speed system using this type of board can tolerate a phase shift in a sine-wave signal due to propagation delay of only \( 15^\circ \) before failing to work. Calculate the maximum length \( l_{\text{MAX}} \) (in meters, or a suitable subdivision such as centimeters or millimeters) that a circuit-board trace carrying this signal can have at each of the following frequencies without encountering excessive phase shift due to the propagation delay along the trace:- (a) \( f = 10 \text{ MHz} \),- (b) \( f = 30 \text{ MHz} \),- (c) \( f = 500 \text{ MHz} \),- (d) \( f = 2.5 \text{ GHz} \).The task is to determine the maximum permissible length of the circuit conductor for each specified frequency to ensure the phase shift does not exceed the tolerable \( 15^\circ \) limit.

Since one wavelength corresponds to the propagation delay of 360 degrees, that implies to a propagation delay of 360 degrees which is corresponding to 20 meters. Hence a propagation delay of 15 degrees corresponding to a distance is Since one wavelength corresponds to the propagation delay of 360 degrees, that implies to a propagation delay of 360 degrees which is corresponding to 6.67 meters. Hence a propagation delay of 15 degrees corresponding to a distance is Since one wavelength corresponds to the propagation delay of 360 degrees, that implies to a propagation delay of 360 degrees which is corresponding to 0.4 meters. Hence a propagation delay of 15 degrees corresponding to a distance is Since one wavelength corresponds to the propagation delay of 360 degrees, that implies to a propagation delay of 360 degrees which is corresponding to 0.08 meters. Hence a propagation delay of 15 degrees corresponding to a distance is