The time range defaults to 1u for one micro second. You must change it to a period appropriate to the signal period you will be using. For example the 1m for one millisecond shown above will allow two full periods of a 2 kHz signal to be analyzed and displayed or one cycle of a 1kHz signal. The display above also shows two waveforms to be displayed on a single graph with time as the x-axis variable. The two y-axis variables are the node voltage at the plus terminal of the voltage source named V1 and the emitter voltage of transistor Q1 the node numbers could also be used, but this illustrates an alternate method of specifying the desired node voltage. Note that both x and y scales have been set to linear. The fifth button in front of each expression allows the specified waveform to be saved to a waveform file to be used later as a user defined waveform source for another circuit. The picture below shows the circuit to be analyzed including the dc voltages found during an ac analysis of the circuit.
Since the Operating Point box has been checked the transient analysis will start with the voltages shown above. That is C1 will be charged to 11.085 volts positive on the base side and C2 will be charged with 10.324 volts plus to minus from emitter to load resistor. as the analysis runs these voltages will be changed as the differential equations describing the complete circuit are solved numerically. The solver will adjust the time step size to try and get a smooth curve while not calculating too many intermediate points. The first run of the simulation is shown below with the amplitude of the sine source set to A=6 giving an input voltage signal of 6sin(2 pi 1000 t) volts. As can be seen this in enough to cause the amplifier to saturate on the first cycle for a short time. It is also enough to cause it to cutoff at the lower portion of the waveform.
See the top waveform distorting and clipping as starting form 10.324 volts the emitter voltage can not go up the full 6 volts of the signal waveform before saturating, since the battery voltage is only 15 volts. Then at the bottom of the wave form it is clipped due to cutoff of the transistor. As the input voltage drops the collector current drops toward zero and the base-emitter junction actually becomes reverse biased as the charge on C2 maintains the emitter slightly above 5 volts. The time constant of the C2-(Re + RL) circuit is 200 ms which allows little change in the voltage during the time of about .136 ms that the transistor is cutoff. The waveform in the first half cycle as it approaches the peak is not smooth and as shown in the next picture is actually off by a very noticeable amount due to the large step size chosen by the program.
The maximum step size can be specified in the analysis limits window. Shown below is the value 0.01 ms used o get the plot above. This limits the program to increasing the time step to no more than 0.01 ms during the calculations. Note the shape of the waveform can also be seen on the output voltage, v(2), shown on this graph. The saturation does not occur on the second cycle because the capacitors have been charged to new voltages by the non symmetrical waveform.
These new voltages are shown below.
Note that now there is almost a full 6 volts between the emitter DC voltage and the collector voltage. There will actually be a short time when the transistor is saturated at the very top of the waveform