Digital Storage Oscilloscope Memory Calculation

A digital storage oscilloscope (DSO) captures and stores electrical signals as digital data. The ability of a DSO to record a transient event, which is a signal that occurs for a short duration, depends heavily on its memory depth and sampling rate. The question asks for the minimum memory size required for a DSO to capture a specific transient event.

Understanding Key Parameters

To determine the necessary memory size, we need to consider two main parameters provided in the question:

• Transient Duration: This is the total length of time the signal event lasts that we want to record. In this case, it is 10 seconds.
• Time Resolution: This refers to how finely the signal is sampled over time. A smaller time resolution means more samples are taken per unit of time, leading to a more detailed capture of the waveform. Here, the time resolution is 200 μs (microseconds).

Calculating Required Samples

The memory size of a digital storage oscilloscope (DSO) is directly related to the number of data points (samples) it can store for a single waveform. To record a transient signal of a certain duration with a specific time resolution, we need to determine how many samples are required over that duration. Each sample corresponds to a data point stored in the oscilloscope's memory.

The formula to calculate the number of samples needed is:

$$ \text{Number of Samples} = \frac{\text{Transient Duration}}{\text{Time Resolution}} $$

Let's substitute the given values into this formula:

• Transient Duration = 10 seconds
• Time Resolution = 200 μs = 200 \(\times\) 10\(^{-6}\) seconds = 0.0002 seconds

Now, perform the calculation:

$$ \text{Number of Samples} = \frac{10 \text{ s}}{200 \times 10^{-6} \text{ s}} $$

$$ \text{Number of Samples} = \frac{10}{0.0002} $$

$$ \text{Number of Samples} = 50,000 $$

This means the digital storage oscilloscope needs to capture and store 50,000 individual data points to record the entire 10-second transient with the specified time resolution.

Determining Memory Size

Each sample typically requires one memory location. Therefore, the total number of samples directly indicates the required memory depth. Since we calculated that 50,000 samples are needed, the memory size of the oscilloscope must be at least 50,000 points.

In digital systems, memory sizes are often expressed using prefixes like 'K' (kilo), 'M' (mega), etc. Here, 'K' usually stands for 1000 (as opposed to 1024 for computer memory in some contexts, but for oscilloscope memory points, 1000 is common).

So, 50,000 samples can be expressed as:

$$ 50,000 \text{ samples} = 50 \times 1000 \text{ samples} = 50 \text{ K samples} $$

Therefore, a memory size of 50 K (50 kilo-samples) is required for the digital storage oscilloscope to meet the given specifications.

The final answer is $\text{50 K}$.